UC-NRLF B M EMfl 7D1 w;iTH:»rst;n':-.!i;;i!MJr Kfl-^I^E; / >ti^^ ^^^^^H&iiRiiiziii. . ^^H ^If>e4 f^^v-it«IMMH£MMt(Mt»ni^ H' ' ^ ^Ki: ^Hibitvii. HHK ^^^^^^H^mBrt ^Hjiy: HIIRh' ^^^!^H||p '^^^^^H^K 1!* 32*b IN MEMORIAM FLORIAN CAJORI 4^i^^a ^ (Sy^ WENTWORTH'S "^ PLANE GEOMETRY REVISED BY GEORGE WENTWORTH AND DAVID EUGENE SMITH GINN AND COMPANY BOSTON ■ NEW YORK • CHICAGO • LONDON COPYRIGHT, 1888, 1899, BY G. A. WENT WORTH COPYRIGHT, 1910, BY GEORGE WENTWORTH AND DAVID EUGENE SMITH ENTERED AT STATIONERS' HALL ALL RIGHTS RESERVED 910.7 GINN AND COMPANY • PRO- PRIETORS • BOSTON • U.S.A. b ^ I^Ai/^ PEEFACE Long after the death of Robert Recorde, England's first great writer of textbooks, the preface of a new edition of one of his works contained the appreciative statement that the book was " entail'd upon the People, ratified and sign'd by the approbation of Time.'' The language of this sentiment sounds quaint, but the noble tribute is as impressive to-day as when first put in print two hundred fifty years ago. With equal truth these words may be applied to the Geome> try written by George A. Wentworth. For a generation it has been the leading textbook on the subject in America. It set a standard for usability that every subsequent writer upon geometry has tried to follow, and the number of pupils who have testified to its excellence has run well into the millions. In undertaking to prepare a revision of this work, the authors have been guided by certain well-defined principles, based upon an extended investigation of the needs of the schools and upon a study of all that is best in the recent literature of the sub- ject. The effects of these principles they feel should be sum- marized for the purpose of calling the attention of the wide circle of friends of the Wentworth text to the points of simi- larity and of difference in the editions. 1. Every effort has. been made not only to preserve but to improve upon the simplicity of treatment, the clearness of expression, and the symmetry of page that have characterized the successive editions of the Wentworth Geometry. It has been the purpose to prepare a book that should do even more than maintain the traditions this work has fostered. rft\£y^ €\9% iv PLANE GEOMETRY 2. The proofs have been given substantially in full, tOv.the end that the pupil may always have before him a model for his independent treatment of the exercises. 3. The sequence of propositions has been improved in sev- eral respects, notably in the treatment of parallels. 4. To meet a general demand, the number of propositions has been decreased so as to include only the great basal theo- rems and problems. A little of the less important material has been placed in the Appendix, to be used or not as circum- stances demand. 5. The exercises, in some respects the most important part of a course in geometry, have been rendered more dignified in appearance and have been improved in content. The number of simple exercises has been greatly increased, while the diffi- cult puzzle is much less in evidence than in most American textbooks. The exercises are systematically grouped, appear- ing in full pages, in large type, at frequent intervals. They are not all intended for one class, but are so numerous as to allow the teacher to make selections from year to year. 6. The introduction has been made as concrete as is reason- able. Definitions have been postponed until they are actually needed, only well-recognized terms have been employed, the pupil is initiated at once into the practical use of the instru- ments, some of the reasons for studying geometry are early shown in an interesting way, and correlation is made with the simple algebra already studied. The authors are indebted to many friends of the Wentworth Geometry for assistance and encouragement in the labor of pre- paring this edition, and they will welcome any further sugges- tions for improvement from any of their readers. GEORGE WENTWORTH DAVID EUGENE SMITH CONTENTS Page INTRODUCTION 1 BOOK I. RECTILINEAR FIGURES 25 Triangles 26 Parallel Lines 46 Quadrilaterals ......... 59 Polygons 68 Loci 73 BOOK II. THE CIRCLE 93 Theorems 94 Problems 126 BOOK III. PROPORTION. SIMILAR POLYGONS . . .151 Theorems 152 Problems 182 BOOK IV. AREAS OF POLYGONS 191 Theorems 192 Problems 214 BOOK V. REGULAR POLYGONS AND CIRCLES . . 227 Theorems 228 Problems 242 APPENDIX 261 Symmetry 261 Maxima and Minima ........ 265 Recreations 273 History of Geometry ........ 277 INDEX 283 V SYMBOLS AND ABBREVIATIONS = equals, equal, equal to, Adj. adjacent. is equal to, or Alt. alternate. is equivalent to. Ax. axiom. > is greater than. Const. construction. < is less than. Cor. corollary. II parallel. Def. definition. _L perpendicular. Ex. exercise. Z angle. Ext. exterior. A triangle. Fig. figure. O parallelogram. Hyp. hypothesis. □ rectangle. Iden. identity. O circle. Int. interior. st. straight. Post. postulate. rt. right. Prob. problem. •.• since. Prop. proposition. .'. therefore. Sup. supplementary. These symbols take the plural form when necessary, as in the case of The symbols +, — , x, -^ are used as in algebra. There is no generally accepted symbol for "is congruent to," and the words are used in this book. Some teachers use = or =, and some use = , but the sign of equality is more commonly employed, the context telling whether equality, equivalence, or congruence is to be understood. Q. E. D. is an abbreviation that has long been used in geometry for the Latin words quod erat demonstrandum, "which was to be proved." Q. E. F. stands for quod erat faciendum, "which was to be done." PLANE GEOMETRY INTRODUCTION 1. The Nature of Arithmetic. In arithmetic we study compu- tation, the working with numbers. We may have a formula expressed in algebraic symbols, such as a = hh, where a may stand for the area of a rectangle, and h and h respectively for the number of units of length in the base and height ; but the actual computation involved in applying such formula to a particular case is part of arithmetic. 2. The Nature of Algebra. In algebra we generalize the arithmetic, and instead of saying that the area of a rectangle with base 4 in. and height 2 in. is 4 x 2 sq. in., we express a general law by saying that a = hh. In arithmetic we may have an equality, like 2 x 16 +17= 49, but in algebra we make much use of equations, like 2 cc -f- 17 = 49. Algebra, therefore, is a generalized arithmetic. 3. The Nature of Geometry. We are now about to begin another branch of mathematics, one not chiefly relating to numbers although it uses numbers, and not primarily devoted to equa- tions although using them, but one that is concerned principally with the study of forms, such as triangles, parallelograms, and circles. Many facts that are stated in arithmetic and algebra are proved in geometry. For example, in geometry it is proved that the square on the hypotenuse of a right triangle equals the sum of the squares on the other two sides, and that the circumference of a circle equals 3.1416 times the diameter, 1 PLANE GEOMETKY 4. Solid. The block here represented is called a solid; it is a limited portion of space filled with matter. In geometry, however, we have nothing to do with the matter of which a body is composed; we study simply its sha^je and size, as in the second figure. That is, a physical solid can be touched and handled ; a geometric solid is the space that a physical solid is conceived to occupy. For example, a stick is a physical solid ; but if we put it into wet plaster, and then remove it, the hole that is left may be thought of as a geometric solid although it is filled with air. 5. Geometric Solid. A limited portion of space is called a geometric solid. 6. Dimensions. The block represented in § 4 extends in three principal directions : (1) From left to right, that is, from A to D] (2) From back to front, that is, from A to B] (3) From top to bottom, that is, from A to E. These extensions are called the dimensions of the block, and are named in the order given, length, breadth (or width), and thickness (height, altitude, or depth). Similarly, we may say that every solid has three dimensions. Very often a solid is of such shape that we cannot point out the length, or distinguish it from the breadth or thickness, as an irregular block of coal. In the case of a round ball, where the length, breadth, and thick- ness are all the same in extent, it is impossible to distinguish one dimen- sion from the others. INTEODUCTIO>T 3 7. Surface. The block shown in § 4 has six flat faces, each of which is called a surface. If the faces are made smooth by polishing, so that when a straight edge is applied to any one of them the straight edge in every part will touch the surface, each face is called 2^ plane surface, or 2^ plane. These surfaces are simply the boundaries of the solid. They have no thickness, even as a colored light shining upon a piece of paper does not make the paper thicker. A board may be planed thinner and thinner, and then sandpapered still thinner, thus coming nearer and nearer to representing what we think of as a geometric plane, but it is always a solid bounded by surfaces. That which has length and breadth without thickness is called a surface. 8. Line. In the solid shown in § 4 we see that two adja- cent surfaces intersect in a line. A line is therefore simply the boundary of a surface, and has neither breadth nor thickness. That which has length without breadth or thickness is called a line. A telegraph wire, for example, is not a line. It is a solid. Even a pencil mark has width and a very little thickness, so that it is also a solid. But if we think of a wire as drawn out so that it becomes finer and finer, it comes nearer and nearer to representing what we think of and speak of as a geometric line. 9. Magnitudes. Solids, surfaces, and lines are called tnag- nitudes. 10. Point. In the solid shown in § 4 we see that when two lines meet they meet in a point. A point is therefore simply the boundary of a line, and has no length, no breadth, and no thickness. That which has only position, without length, breadth, or thickness, is called di: point. We may think of the extremity of a line as a point. We may also think of the intersection of two lines as a point, and of the intersection of two surfaces as a line. 4 PLANE GEOMETRY 11. Representing Points and Geometric Magnitudes. Although we only imagine such geometric magnitudes as lines or planes, we may represent them by pictures. Thus we represent a point by a fine dot, and I \ 7 name it by a letter, as P in this figure. / / We represent a hne by a fine mark, and name L ' it by letters placed at the ends, as ^5. We represent a surface by its boundary lines, and name it by letters placed at the corners or in some other convenient way, as A BCD. We represent a solid by the boundary faces or by the lines bounding the faces, as in § 4. 12. Generation of Geometric Magnitudes. We may think of (1) A line as generated by a moving point ; (2) A surface as generated by a moving line ; (3) A solid as generated by a moving surface. For example, as shown in the figure let the surface A BCD move to the position WXYZ. Then (1) A generates the line AW; (2) AB generates the surface A WXB ; D (3) ABCD generates the solid AY. C ., Y A — >— B 71 ->■ Of course a point will not generate a line / ' / by simply turning over, for this is not mo- ^ ^ ^ tion for a point ; nor will a line generate a surface by simply sliding along itself ; nor will a surface generate a solid by simply sliding upon itself. 13. Geometric Figure. A point, a line, a surface, a solid, or any combination of these, is called a geometric figure. A geometric figure is generally called simply a figure. 14. Geometry. The science of geometric figures is called geometry. Plane geometry treats of figures that lie wholly in the same plane, that is, of plane figures. Solid geometry treats of figures that do not lie wholly in the same plane. INTEODUCTIOK 15. Straight Line. A line such that any part placed with its ends on any other part must lie wholly in the line is called a straight line. For example, J.i> is a straight line, for if we take, say, a half inch of it, and place it in any way on any other part of AB, but so that its ends lie in AB^ then the whole of the half inch of line will lie in AB. This is well shown by using tracing paper. The word line used alone is understood to mean a straight line. Part of a straight line is called a segment of the line. The term seg- ment is applied also to certain other magnitudes. 16. Equality of Lines. Two straight-line segments that can be placed one upon the other so that their extremities coin- cide are said to be equal. In general, two geometric magnitudes are equal if they can be made to coincide throughout their whole extent. We shall see later that some figures that coincide are said to be congruent. 17. Broken Line. A line made up of two or more different straight lines is called a broken line. For example, CD is a broken line. 18. Rectilinear Figure. A plane figure bounded by a broken line is called a rec- tilinear figure. For example, A BCD is a rectilinear figure. 19. Curve Line. A line no part of which is straight is called a curiae line, or simply a curve. For example, EF is a curve line. 20. Curvilinear Figure. A plane figure formed by a curve line is called a curvilinear figure. For example, is a curvilinear figure with which we are already familiar. Some curvilinear figures are surfaces bounded by curves and others are the curves themselves. •O 6 PLANE GEOMETEY 21. Angle. The opening between two straight lines drawn from the same point is called an angle. Strictly speaking, this is a plane angle. We shall find later that there are angles made by curve lines and angles made by planes. The two lines are called the sides of the angle, and q the point of meeting is called the vertex. An angle may be read by naming the letters desig- nating the sides, the vertex letter being between the others, as the angle A OB. An angle may also be desig- nated by the vertex letter, as the angle O, or by a small letter within, as the angle m. A curve is often drawn to show the par- ticular angle meant, as in angle m. 22. Size of Angle. The size of an angle depends upon the amount of turning necessary to bring one side into the position of the other. One angle is greater than another angle when the amount of turning is greater. Thus in these compasses the first angle is smaller than the second, which is also smaller than the third, length of the sides has nothing to do with the size of the angle. The 23. Equality of Angles. Two angles that can be placed one upon the other so that their vertices coincide and the sides of one lie along the sides of the other are said to ^b be equal. For example, the angles AOB and A'O'B' (read "J. prime, prime, B prime") are equal. It is well to illustrate this by tracing one on thin paper and placing it upon the other. q'^c ^ 24. Bisector. A point, a line, or a plane that divides a geo- metric magnitude into two equal parts is called a bisector of the magnitude. For example, M, the mid-point of the line AB^ A M B is a bisector of the line. INTRODUCTION 25. Adjacent Angles. Two angles that have the same vertex and a common side between them are called adjacent angles. For example, the angles AOB and BOC are adjacent angles, and in §26 the angles AOB and BOC are adjacent angles. 26. Right Angle. When one straight line meets another straight line and makes the adjacent angles equal, each angle is called a right angle. For examplie, angles A OB and BOCm this figure. If CO is cut off, angle AOB is still a right angle. ^ ^ ^ 27. Perpendicular. A straight line making a right angle with another straight line is said to be ijeirpendicular to it. Thus OB is perpendicular to CA^ and CA to OB. OB is also called a perpendicular to CJ., and is called the foot of the perpendicular OB. 28. Triangle. A portion of a plane bounded by three straight lines is called a triangle. C The lines AB, BC, and CA are called the sides of the triangle ABC^ and the sides taken together form the perimeter. The points A, B, and C are the vertices of the triangle, and the angles A, B, and C are the angles of the triangle. The side AB upon which the triangle is supposed to rest is the base of the triangle. Similarly for other plane figures. 29. Circle. A closed curve lying in a plane, and such that all of its points are equally distant from a fixed point in the plane, is called a circle. The length of the circle is called the circumference. The point from which all points on the circle are equally distant is the center. Any portion of a circle is an arc. A straight line from the center to the circle is a radius. A straight line through the center, termi- nated at each end by the circle, is a diameter. Formerly in elementary geometry circle was taken to mean the space inclosed, and the bounding line was called the circumference. Modern usage has conformed to the definition used in higher mathematics. 8 PLANE GEOMETRY 30. Instruments of Geometry. In geometry only two instru- ments are necessary besides pencil and paper. These are a straight edge, or ruler, and a pair of compasses. It is evident that all radii of the same circle are equal. In the absence of compasses, and particularly for blackboard work, a loop made of string may be used. For the accurate transfer of lengths, however, compasses are desirable. 31. Exercises in using Instruments. The following simple exercises are designed to accustom the pupil to the use of instruments. No proofs are attempted, these coming later in the course. This section may be omitted if desired, without affecting the course. *.CL EXERCISE 1 1. From a given point on a given straight line required to draw a perpendicular to the line. Let AB he the given line and P be the given point. It is required to draw from P a line per- pendicular to AB. With P as a center and any convenient radius draw arcs cutting AB 2it X and Y. With JT as a center and XY as a radius draw a circle, and with Y as a center and the same radius draw another circle, and call one inter- section of the circles C. With a straight edge draw a line from P to C, and this will be the perpendicular required. \X Yl INTRODUCTION 9 A X^' 2. From a given point outside a given straight line required to let fall a perpendicular to the line. p Let AB he the given straight line and P be the given point. It is required to draw from P a line perpen- dicular to AB. With P as a center and any convenient radius draw an arc cutting AB at X and Y. With X as a center and any convenient radius draw a circle, and with F as a center and the same radius draw another circle, and call one intersection of the circles C. With a straight edge draw a straight line from P to C, and this will be the perpendicular required. It is interesting to test the results in Exs. 1 and 2, by cutting the paper and fitting the angles together. -S^' I 3. Required to draw a triangle having two sides each equal to a given line. Let I be the given line. It is required to draw a triangle having two sides each equal to I. With any center, as C, and a radius equal to I draw an arc. Join any two points on the arc, as A and J5, with each other and with C by straight lines. Then ABC is the triangle required. ^'-~ 4. Required to draw a triangle having its three sides each equal to a given line. ^ ^ ^ Let AB he the given line. It is required to draw a triangle having its three sides each equal to AB. With A as a center and ^Z? as a radius draw a circle, and with B as a center and the same radius draw another circle. Join either intersection of the circles with A and B by straight lines. Then ABC is the triangle required. In such cases draw the arcs only long enough to show the point of intersection. 10 PLANE GEOMETPvY ^ ^s.O.'-- 5. Eequired to draw a triangle having its sides equal respec- tively to three given lines. Let the three lines be I, m, and n. What is now required ? Upon any line mark off with the com- passes a line-segment AB equal to I. With ^ as a center and m as a radius draw a circle ; with 2^ as a center and n as a radius draw a circle. Drawee and 50. ^ Then ABC is the required triangle. n 6. From a given point on a given line required to draw a line making an angle equal to a given angle. X ^^-- o \c IM Let P be the given point on the given line PQ, and let angle AOB be the given angle. What is now required ? With as a center and any radius draw an arc cutting ^ O at C and BO at D. With P as a center and OC as a radius draw an arc cutting PQ at M. With Jlf as a center and the straight line joining C and D as a radius draw an arc cutting the arc just drawn at N, and draw PN. Then angle MPN is the required angle. 7. Eequired to bisect a given straight line. Let AB be the given line. It is required to bisect it. With J. as a center and ^B as a radius draw a circle, and with B as a center and the same radius draw a circle. Call the two intersections of the circles X and Y. Draw the straight line XY. Then XY bisects the line AB a^t the point of inter- section M. ^C M -r INTKODUCTION 11 8. Required to bisect a given angle. Let A OB be the given angle. It is required to bisect it. With O as a center and any convenient radius draw an arc cutting OA at X and OB at Y. With X as a center and a line joining X and F as a radius draw a circle, and with F as a center and the same radius draw a circle, and call one point of inter- section of the circles P. Draw the straight line OP. Then OP is the required bisector. 9. By the use of compasses and ruler draw the following figures : ^^ ^ The dotted lines show how to fix the points needed in drawing the figure, and they may be erased after the figure is completed. In general, in geometry, auxiliary lines (those needed only as aids) are indicated by dotted lines. 10. By the use of compasses and ruler draw the following figures : It is apparent from the figures in Exs. 9 and 10 that the radius of the circle may be used in describing arcs that shall divide the circle into six equal parts. 12 PLANE GEOMETRY 11. By the use of compasses and ruler draw the following figures : 12. By the use of compasses and ruler draw the following figures : 13. By the use of compasses and ruler draw the following figures : In such figures artistic patterns may be made by coloring various portions of the drawings. In this way designs are made for stained-glass windows, for oilcloth, for colored tiles, and for other decorations. 14. Draw a triangle of which each side is li in. 15. Draw two lines bisecting each other at right angles. INTRODUCTION 13 16. Bisect each of the four right angles formed by two lines bisecting each other at right angles. 17. Draw a line 1^- in. long and divide it into eighths of an inch, using the ruler. Then with the compasses draw this figure. It is easily shown, when we come to the measurement of the circle, that these two curve lines divide the space inclosed by the circle into parts that are exactly equal to one another. By continuing each semicircle to make a complete circle another inter- esting figure is formed. Other similar designs are easily invented, and students should be encouraged to make such original designs. 18. In planning a Gothic window this drawing is needed. The arc BC is drawn with ^ as a center q and vlJ5 as a radius. The small arches are described with A, Z), and B as centers and ^i) as a radius. The center P is found by taking A and B as centers and AE sls a l-adius. How may the points D, E, and F be found ? Draw the figure. j_ p ^ E B 19. Draw a triangle of which each side is 1 in. Bisect each side, and with the points of bisection as centers and with radii ^ in. long draw three circles. 20. A baseball diamond is a square 90 ft. on a side. Draw the plan, using a scale of ^^ in. to a foot. Locate the pitcher 60 ft. from the home plate. 21. A man travels from A directly east 1 mi. to B. He then turns and travels directly north 1| mi. to C. Draw the plan and find by measurement the distance ^ C to the nearest quarter of a mile. Use a scale of ^ in. to a mile. 14 PLANE GEOMETRY 22. A double tennis court is 78 ft. long and 36 ft. wide. The net is placed 39 ft. from each end and the service lines 18 ft. from each end. Draw the plan, using a scale of y^^ in. to a foot, making the right angles as shown in Ex. 1. The accuracy of the construction may be tested by measuring the diagonals, which should be equal. 23. At the entrance to New York harbor is a gun having a range of 12 mi. Draw a line inclosing the range of fire, using a scale of ^^ in. to a mile. 24. Two forts are placed on opposite sides of a harbor entrance, 13 mi. apart. Each has a gun having a range of 10 mi. Draw a -plan showing the area exposed to the fire of both guns, using a scale of ^\ in. to a mile. 25. Two forts, A and B, are placed on opposite sides of a harbor entrance, 16 mi. apart. On an island in the harbor, 12 mi. from A and 11 mi. from B, is a fort C. The fort A has a gun with a range of 12 mi., fort B one with a range of 11 mi., and fort C one with a range of 10 mi. Draw a plan of the entrance to the harbor, showing the area exposed to the fire of each gun. 26. A horse, tied by a rope 25 ft. long at the corner of a lot 50 ft. square, grazes over as much of the lot as possible. The next day he is tied at the next corner, the third day at the third corner, and the fourth day at the fourth corner. Draw a plan showing the area over which he has grazed during the four days, using a scale of ^ in. to 5 ft. 27. A gardener laid out a flower bed on the following plan : He made a triangle ABC, 16 ft. on a side, and then bisected two of the angles. From the point of intersection of the bi- sectors, P, he drew perpendiculars to the three sides of the triangle, PX, PY, and PZ. Then he drew a circle with P as a center and PX as a radius, and found that it just fitted in the triangle. Draw the plan, using a scale of i in. to a foot. IKTEODUCTIOK 15 32. Necessity for Proof. Although pUrt of geometry consists in drawing figures, this is not the most important part. It is essential to prove that the figures are what we claim them to be. The danger of trusting to appearances is seen in Exercise 2. EXERCISE 2 1. Estimate which is the longer line, AB or XF, and how much longer. Then test your estimate by ^> <^B measuring with the compasses or with a piece of paper carefully marked. 2. Estimate which is the longer line, AB ot CDj and how much longer. Then test your estimate by measuring as in Ex. 1. 3- Look at this figure and state whether AB and CD are both straight lines. If one is not straight, which one is it? Test your answer by us- ing a ruler or the folded edge of a piece of paper. 4. Look at this figure and state whether AB and CD are the same distance apart at .4 and C as at B and D. Then test your answer as in Ex. 1. 5 . Look at this figure and state whether AB will, if prolonged, lie on CD. Also state whether WX will, if prolonged, lie on YZ. Then test your answer <- ^ mmm^} ^ by laying a ruler along the lines. ^^ 6. Look at this figure and state which of the three lower lines is ^-B prolonged. Then test your answer by laying a ruler along AB, i 16 PLANE GEOMETRY 33. Straight Angle. When the sides of an angle extend in opposite directions, so as to be in the same straight line, the angle is called a straight angle. ^-^ For example, the angle A OB, as shown in this ^ figure, is a straight angle. The angle BOA, below the line, is also a straight angle. 34. Right Angle and Straight Angle. It follows from the definition of right angle (§ 26) that a right angle is half of a straight angle. In like manner, it follows that a straight angle equals twice a right angle. 35. Acute Angle. An angle less than a right angle is called an aciite angle. For example, the angle m, as shown in this figure, is an acute angle. 36. Obtuse Angle. An angle greater than a right angle and less than a straight angle is called an obtuse „ angle. For example, the angle AOB, as shown in this . , figure, is an obtuse angle. ''^9^y 37. Reflex Angle. An angle greater than a straight angle and less than two straight angles is called a reflex angle. For example, the angle BOA, marked with a dotted curve line in the figure in § 36, is a reflex angle. When we speak of an angle formed by two given lines drawn from a point we mean the smaller angle unless the contrary is stated. 38. Oblique Angles. Acute angles and obtuse angles are called oblique angles. The sides of oblique angles are said to be oblique to each other, and are called oblique lines. Evidently if we bisect a straight angle, we form two right angles ; if we bisect a right angle or an obtuse angle, we form two acute angles ; if we bisect a reflex angle, we form two obtuse angles. INTEODUCTION 17 39. Generation of Angles. Suppose the line r to revolve from the position OA about the point O as a vertex to the posi- tion OB. Then r describes or generates the acute angle A OB, and, as we have seen (§ 22) the size of the angle depends upon the amount of rotation, the angle being greater as the amount of turning is greater. If r rotates still further, to the position OC, it has then generated the right angle AOC and is perpendicular to OA. If r rotates still further, to the position OD, it has then generated the obtuse angle A OB. If r rotates to the position OE^ it has then generated the straight angle AOE. If r rotates to the position OF, it has then generated the reflex angle A OF. If r rotates still further, past 06? to the position OA again, it has made a complete revolution and has generated two straight angles or four right angles. 40. Sums and Differences of Magnitudes. If the straight line AP has been generated by a point P , i___+__>P moving from A to P, the segments ^ t B CD AB, BC, CD, and so on, having been generated in succession, then we call AC the stem oi AB and BC. That is, AC = AB-\- BC, whence AC — BC = AB. If the angle A OD has been generated by the - (^ line OA revolving about as a vertex from the position OA, the angles AOB, BOC, and COD having been generated in succession, then we call angle AOC the sum of angles AOB and BOC. That is, considering angles, .4 OC = ^ 0^ + BOC, whence AOC - BOC = A OB. In the same way that we may have the sum or the difference of lines or of angles we may have the sum or the difference of surfaces or of solids. 18 PLANE GEOMETRY 41. Perigon. The whole angular space in a plane about a point is called a perigon. It therefore follows that a perigon equals the sum of two straight angles or the sum of four right angles. 42. Complements, Supplements, and Conjugates, If the sum of two angles is a right angle, each angle is called the comple- ment of the other. If the sum of two angles is a straight angle, each angle is called the supplement of the other. If the sum of two angles is a perigon, each angle is called the conjugate of the other. Thus, with respect to angle AOB, the complement is angle BOC, the supplement is angle J50D, the conjugate is angle BOA (reflex). 43. Properties of Supplementary Angles. It is sufficiently evi- dent to be taken without proof that 1. The two adjacent angles which one straight line Tuakes with another are together equal to a straight angle. 2. If the sum of two adjacent angles is a straight angle, their exterior sides are in the sam,e straight line. 44. Angle Measure. Angles are measured by taking as a unit •j^-Q of a perigon. This unit is called a degree. The degree is divided into 60 equal parts, called minutes, and the minute into 60 equal parts, called seconds. We write 5° 13' 12" for 5 degrees 13 minutes 12 seconds. It is evident that a right angle equals 90°, a straight angle equals 180°, and a perigon equals 360°. 45. Vertical Angles. When two angles have the same vertex, and the sides of the one are prolongations of the sides of the other, those angles are called \ vertical angles. ^\" In the figure the angles x and z are vertical angles, \ as are also the angles w and y. \ INTRODUCTION 19 EXERCISE 3 1. Find the complement of 72°; of 65^30'; of 22° 20' 15". 2. What is the supplement of 45° ? of 120° ? of 145° 5' ? of 22° 20' 15" ? 3. What is the conjugate of 240° ? of 280° ? of 312° 10' 40" ? 4. The complement of a certain angle x is 2x. How many degrees are there in x ? yx 5. The complement of a certain angle ic is 3ic. How many degrees are there in a? ? 6. What is the angle of which the complement is four times the angle itself ? 7. The supplement of a certain angle a; is 5 ic. ^^ How many degrees are there in ic ? ^xy^ 8. The supplement of a certain angle x is 14 x. How many degrees are there in cc ? 9. What is the angle of which the supplement equals half of the angle itself ? 10. How many degrees in an ahgle that equals its own com- plement ? in one that equals its own supplement ? 11. The conjugate of a certain angle ic is fee. A ^ A How many degrees are there in cc ? y^ 12. The conjugate of a certain angle x is \x. How many degrees are there in cc ? 13. How many degrees in an angle that equals a third of its own conjugate? in one that equals its own conjugate? 14. Find two angles, x and y, such that their sum is 90° and their (difference is 10°. 15. Find two complementary angles such that their differ- ence is 30°. 16. Find two supplementary angles such that one is 20° greater than the other. 20 PLANE GEOMETRY 17. The angles x and y are conjugate angles, and their differ- ence is a straight angle. How many degrees are there in each ? 18. The angles x and y are conjugate angles, and their differ- ence is zero. How many degrees are there in each ? 19. Of two complementary angles one is four fifths of the other. How many degrees are there in each ? 20. Of two supplementary angles one is five times the other. How many degrees are there in each ? 21. How many degrees are there in the smaller angle formed by the hands of a clock at 5 o'clock ? 22. How many degrees are there in the smaller angle formed by the hands of a clock at 10 o'clock ? b 23. In this figure, if angle A OB is 38°, how y^ many degrees in angle BOC ? How many in Xo angle COD? How many in angle DO A ? j{ 24. In the same figure, if angle AOB is equal to a third of angle BOC, how many degrees in each of the four angles ? 25. In the angles of this figure, \iw = 2x, how many degrees in each ? How many degrees iny? \ How many degrees in ^ ? A^ 26. Find the angle whose complement de- \ creased by 30° equals the angle itself. 27. Find the angle whose complement divided by 2 equals the angle itself. 28. Draw a figure to show that if two adjacent angles have their exterior sides in the same straight line, their sum is a straight angle. 29. Draw a figure to show that the sum of all the angles on the same side of a straight line, at a given point, is equal to two right angles. 30. Draw a figure to show that the complements of equal angles are equal. INTRODUCTION 21 46. Axiom. A general statement admitted without proof to be true is called an axiom. For example, it is stated in algebra that "if equals are added to equals the sums are equal." This is so simple that it is generally accepted without proof. It is therefore an axiom. 47. Postulate. In geometry a geometric statement admitted without proof to be true is called a postulate. For example, it is so evident that all straight angles are equal, that this statement is a postulate. It is also evident that a straight line may be drawn and that a circle may be described, and these statements are therefore postulates of geometry. Axioms are therefore general mathematical assumptions, while geo- metric postulates are the assumptions peculiar to geometry. Postulates and axioms are the assumptions upon which the whole science of mathe- matics rests. 48. Theorem. A statement to be proved is called a theorem. For example, it is stated in arithmetic that the square on the hypote- nuse of a right triangle equals the sum of the squares on the other two sides. This statement is a theorem to be proved in geometiy. 49. Problem. A construction to be made so that it shall satisfy certain given conditions is called a prohlem. For example, required to construct a triangle all of whose sides shall be equal. This construction was made in § 31, Ex. 4, and later it will be proved that the construction was correct. 50. Proposition. A statement of a theorem to be proved or a problem to be solved is called a proposition. In geometry, therefore, a proposition is either a theorem or a problem. We shall find that most of the propositions at first are theorems. After we have proved a number of theorems so that we can prove that the solutions of problems are correct, we shall solve some problems. 51. Corollary. A truth that follows from another with little or no proof is called 2, corollary. For example, since we admit that all straight angles are equal, it follows as a corollary that all right angles are equal, since a right angle is half of a straight angle. 22 PLANE GEOMETRY 52. Axioms. The following are the most important axioms used in geometry: 1. If equals are added to equals the sums are equal. 2. If equals are subtracted from equals the remainders are equal. 3. If equals are m^ultvplied by equals the products are equal. 4. If equals are divided by equals the quotients are equal. In division the divisor is never zero, 5. Like powers or like p)ositive roots of equals are equal. We learn from algebra that the square root of 4 is + 2 or — 2, but of course these are not equal. In geometry we shall use only the positive roots. 6. If unequals are operated on by positive equals in the same way, the results are unequal in the same order. Taking a>h and taking x and y as equal positive quantities, this axiom states that a4-x>& + 2/, a — x>b — y, ax>by, ->-, etc. X y 7. If unequals are added to unequals in the same order, the sum,s are unequal in the same order ; if unequals are subtracted from equals the remainders are unequal in the reverse order. If a > 6, c > d, and x = y, then a + c>h -^ d, and x — a <y — h. 8. Quantities that are equal to the same quantity or to equal quantities are equal to each other. 9. A quantity may be substituted for its equal in an equation or in an inequality. Thus if X = & and \i a -{- x = c, then a + 6 = c ; and if a + x > c, then a + 6 > c. Axiom 8 is used so often that it is stated separately, although it is really included in Axiom 9. 10. If the first of three quantities is greater than the second, and the second is greater than the third, then the first is greater than the third. Thus if a > 6, and if 6 > c, then a> c. 11. The whole is greater than any of its parts, and is equal to the sum of all of its pjarts. INTEODUCTION 23 53. Postulates. The following are among the most impor- tant postulates used in geometry. Others will be . introduced as needed. 1. One straight line and only one can hfi drawn through two given points. 2. A straight line may be produced to any required length. To produce A B means to extend it through B ; to produce BA means to extend it through A. 3. A straight line is the shortest path between two p)oints. 4. A circle may be described with any given j^oint as a center and any given line as a radices. 5. Any figure may be 7noved from one 2jl(^(^(i to another with- out altering its size or shajje. 6. All straight angles are equal. 54. Corollary 1. Ttvo p>oints determine a straight line. This is only a brief way of stating Postulate 1. 55. Corollary 2. Two straight lines can intersect in only one point. For if they had two points in common they would coincide (Post. 1). 56. Corollary 3. All right angles are equal. For all straight angles are equal (Post. 6), and a straight angle (§ 34) is twice a right angle. Hence Axiom 4 applies. 57. Corollary 4. From a given point in a given line only one perpendicular can be drawn to the line. C b For if there could be two perpendiculars to DA at 0, as OB and OC, we should have angles AOB and AOC both right angles, which is impossible (§ 56). j) 58. Corollary 5. Equal angles have equal complements, equal supplements, and equal conjugates. 59. Corollary 6. The greater of two angles has the less complement, the less suj^plcTnent, and the less conjugate. 24 PLANE GEOMETRY EXERCISE 4 1. If 10° + Z ic = 27° 30', find the value of Z x. 2. If Z a; + 37° = I Z ic + 40°, find the value of Z x. 3. If |Za: + Z^> = 5Z^, find the value of Zic. 4. lfZ.x-\-/-a = ^/.a — Z.X, find the value of Z ic. ^mc? the value of /.x in each of the following equations : 5. Za;+13° = 39°. 10. Zic = 0.7Z;i; + 33°. 6. Zee -17° = 46°. 11. Za^ = 0.1Zic+18°. ^ 7. 2 Zee -ZicH- 23°. 12. fZec = ^Zec + 2^°. 8. 5Zx = 2Zx-4-21°. 13. | Zee = 0.1 Zee +14°. 9. 4Zee = iZee+70°. 14. f Za^ = JZee + 2°. 15. 12Za^+17°=9Z.r + 32°. 16. 5Za;-22°30' = 2Zec+ll°. 17. 51°20'-§Z:r = 5°l' + 3Za;. 18. 73° 21' 4" - Z X = 3° 3' 12" + 4 Z ee. 19. If ec + 20° = y and ?/ — 5° = 2 ee, what is the value of ec and of ?/ ? Find the value of x and of y in each of the following sets of equations : 20. ee + ?/ = 45°, 23. ^ + 2^ = 21°, x-tj = S5°. ^ + 32/ = 26°15'. 21. x-Stj = 0'', 24. ee + 2/ = 9° 20' 15", a; + 8?/=:80°. 2ee-2/ = 12°25' 15". 22. 2ee + 2/ = 64°, 25. ee - 2/ = 5'5", See - 2/ = 88°. 3ee + 4 1/ = 14° 50' 50". 26. If ec < 10° and y = 7° 30', what can be said as to the value oi x-i-y? 27. In Ex. 26, what can be said as to the value oi x — y? BOOK I RECTILINEAR FIGURES Proposition I. Theorem 60. If tioo lines intersect, the vertical angles are equal. D^ A Given the lines AC and BD intersecting at O. To prove that AAOB = Z.COD. Proof. Z .1 OB-\-A BOC = a st. Z. § 43 ( The two adjacent angles which one straight line makes with another are together equal to a straight angle.) Likewise Z50C -|-Z COT) = a st. Z. §43 .-.ZAOB + Z BOC = Z BOC -\- Z COD. Post. 6 (All straight angles are equal.) .\ZAOB = ZCOD. Ax. 2 (If equals are subtracted from equals the remainders are equal.) Q.E.D. 61. Nature of a Proof. From Prop. I it is seen that a tlieorem has (1) certain things given; (2) a definite thing to he p)roved ; (3) a proof, consisting of definite statements, each supported by the authority of a definition, an axiom, a postulate, or some proposition previously proved. 25 26 BOOK I. PLANE GEOMETEY 62. Triangles classified as to Sides. A triangle is said to be scalene when no two of its sides are equal ; isosceles when two of its sides are equal ; equilateral when all of its sides are equal. Scalene Isosceles Equilateral 63. Triangles classified as to Angles. A triangle is said to be right when one of its angles is a right angle ; obtuse when one of its angles is an obtuse angle ; acute when all of its angles are acute angles ; equiangular when all of its angles are equal. Right Obtuse Acute Equiangular 64. Corresponding Angles and Sides. If two triangles have the angles of the one respectively equal to the angles of the other, the equal angles are called corresponding angles, and the sides opposite these angles are called corresponding sides. Corresponding parts are also called homologous parts. 65. Square. A rectilinear figure having four equal sides and four right angles is called a square. 66. Congruent. If two figures can be made to coincide in all their parts, they are said to be congruent. 67. Corollary. Corresponding parts of congruent figures are equal. When equal figures are necessarily congruent, as in the case of angles or straight lines, the word equal is used. For symbols see page vi. TKIANGLES 27 Proposition II. Theorem 68. Tivo triangles are congruent if tivo sides and the included angle of the one are equal respectively to tivo sides and the included angle of the other. Given the triangles ABC and XYZ, with AB equal to XYy AC equal to XZ, and the angle A equal to the angle X To prove that A ABC is congruent to AXYZ. Proof. Place the A ABC upon the AXYZ so that A shall fall on X and AB shall fall along AT. Post. 5 {Any figure may be moved from one place to another without altering its size or shape.) Then B will fall on F, {For AB is given equal to XY.) ^C will fall along XZ, {For ZA is given equal to ZX.) and C will fall on Z. {For AC is given equal to XZ.) .'. CB will coincide with ZY. Post. 1 {One straight line and only one can he drawn through two given points.) .'. the two A coincide and are congruent, by § 66. q.e.d. 69. Corollary. Two right triangles are congruent if the sides of the right angles are equal respectively. The right angles are equal {§ 56). How does Prop. II apply ? 28 BOOK I. PLANE GEOMETEY EXERCISE 5 1. In this figure if Z a = 53°, how many degrees are there inZ^/? inZcc? inZ^? 2. In Ex. 1, if Z a were increased to 89°, what ^^^ would then be the size of A x, y, and z ? 3. In the square ABCD, prove that ^C = J5Z). In ^ABC and BAB what two sides of the one are known to be equal to what two sides of the other ? How about the included angles ? "Write a complete proof as in Prop. II. 4. If ABCD is a square and P is the mid- point of AB, prove that PC = PD. What triangles should be proved congruent ? Can this be done by Prop. II ? Write the proof. 5. How many degrees in an angle that equals a . p b one fourth of its complement ? one tenth of its complement ? 6. How many degrees in an angle that equals twice its supplement ? one third of its supplement ? 7. In the square ABCD the points P, Q, i?, 5 bisect the consecutive sides. Prove that PQ=::QR = RS=SP. 8. In the square ABCD the point P bisects CD, and BM is made equal to AN, as shown in this figure. Prove that PM = PN. What two sides and included angle of one triangle must be proved equal to what two sides and included N angle of another triangle ? 9. Prove that to determine the distance AB across a pond one may sight from A across a post P, place a stake at A' making PA' = AP, then sight along BP making PB' = BP, and finally measure A'B'. ^^ TRIANGLES 29 70. Drawing the Figures. Directions have already been given (§ 31) for drawing the most common geometric figures. For example, in Prop. II the complete work of drawing AXYZ so that XY=AB, Z.X = /.A, and XZ = AC, is indicated in the following figures, the construction lines being dotted^ as is always the case in this book. It is desirable to construct such figures accurately, employing com- passes and ruler until such time as the use of these instruments is thoroughly understood. Eventually, however, the figures should be rapidly but neatly draw^n, free-hand or with the aid of the ruler, as the mathematician usually makes his figures. 71. Designating Corresponding Sides and Angles. It is helpful in propositions concerning equality of figures to check the equal parts so that the eye can follow the proof more easily. Thus it would be convenient to represent the above figures as follows : C Z A • B X Y Here AB and XFhave one check, AC and XZ two checks, and the equal angles A and X are marked by curved arrows. If a figure is very complicated, there is sometimes an advan- tage in using colored crayons or colored pencils, but otherwise this expedient is of little value. While such figures have some attraction for the eye they are not gen- erally used in practice, one reason being that the student rarely has a supply of colored pencils at hand when studying by himself. 80 BOOK I. PLANE GEOMETRY Proposition III. Theorem 72. Two triangles are congruent if two angles and the included side of the one are equal respectively to tivo angles and the included side of the other. T Given the triangles ABC and XYZ, with angle A equal to angle X, angle B equal to angle F, and with AB equal to XY. To prove that A ABC is congruent to AXYZ. Proof. Place the A ABC upon the AXYZ so that AB shall coincide with its equal, XY. Post. 5 {Any figure may be moved from one place to another without altering its size or shape.) Then AC will fall along XZ and BC along YZ. {For it is given that ZA = ZX and ZB = ZY.) .'. C will fall on Z. § 55 {Two straight lines can intersect in only one point.) .'. the two A are congruent. § 66 {If two figures can he made to coincide in all their parts, they are said to be congruent.) Q.E.D. 73. Hypothesis. A supposition made in an argument is called an h7/j)othesls. Thus, where it is said that ZA = ZX and ZB = ZY, yve might say- that this is true "by hypothesis," instead of saying that Z^ is given equal to Z X, and Z jB is given equal to Z Y. The word is generally used, however, for an assumption made somewhere in the proof. TRIANGLES 31 M EXERCISE 6 1. In the square ABCD the point 7^ bisects CD, and PQ and PR are drawn so that Z QPC = 30° and Z RPQ = 120°. Prove that PQ = PR. If ZQPC = 30° and ZRPQ = 120°, what does ZDPR equal ? In the two triangles what parts are respectively- equal, and why ? Write the proof in full. A 2. In this figure prove that if CAT bisects Z .1 CB and is also perpendicular to AB, the triangle ABC is isosceles. ^ In i^AMC and BMC are two angles of the one respec- tively equal to two angles of the other ? Why ? The two triangles have one common side. Write the proof in full. 3. In the triangle ABC, AC = BC and CM bisects the angle C. Prove that CM bisects the base AB. 4. The triangle ABC has Z.A equal to Z.B. The point P bisects AB, and the lines PM and PN are drawn so that Z.BPM=Z.NPA. Prove t\\2it BM = AN. 5. The triangle ABC has Z .1 =Z.B. The lines AP and BQ are so drawn that ZBAP = Z.QBA. Prove that AP = BQ. 6. Wishing to measure the distance across a river, some boys sighted from A to a point P. p They then turned and measured AB at right angles to AP. They placed a stake at 0, halfway from A to B, and drew a perpendicular to AB at B. They placed a stake at C, on this perpendicular, and in line with and P. They then found the width by measuring BC. Prove that they were right 32 BOOK I. PLANE GEOMETRY Proposition IV. Theorem 74. In an isosceles triangle the angles opposite the equal sides are equal. c Given the isosceles triangle ABC^ with AC equal to BC. To prove that Z.A = /.B. Proof. Suppose CD drawn so as to bisect Z A CB. Then in the A ADC and BDC, AC = BC, Given CD = CD, Iden. {That is, CD is common to the two triangles.) and ZACD = ZDCB. Hyp. {For CD bisects Z AC B.) .-.A ADC is congruent to A BDC. § 68 ( Two A are congruent if two sides and the included Z of the one are equal respectively to two sides and the included Z of the other.) ,\ZA=ZB. §67 {Corresponding parts of congruent figures are equal.) Q. E. D. This proposition has long been known as the Pons asinorum, or Bridge of Fools (asses). It is attributed to Thales, a Greek philosopher. In an isosceles triangle the side which is not one of the two equal sides is called the base. 75. Corollary. An equilateral triangle is equiangular. Is ah equilateral triangle a special kind of isosceles triangle ? tria:kgles 33 EXERCISE 7 1. With the figure of Prop. IV, if AC = BC and CD bisects Z C, prove that CD is ± to AB. What angles must be proved to be right angles ? What is a right angle ? Do these angles fulfill the require- ments of the definition ? 2. In the adjacent figure ^C = -BC. Prove that ^ /-in=^ Z. n. 3. In the following figure AC = BC and AD = BD. Prove that Z.CBD=^ADAC. What angles are equal by Prop. IV ? Then what axiom applies ? 4. In the figure of Ex. 3 prove that if a line ^ is drawn from C to D, the A DBC is congruent to the A DA C. b 5. Two isosceles triangles, ABC and ABD, are constructed on the same side of the common base AB. Prove thsA. Z.CBD = Z.DAC. c 6. In the figure of Ex. 5 prove that a line drawn through C and D bisects Z ADB. What two triangles must be proved congruent ? 7. Inthisfigure.4C = 5Cand^P=BQ. Prove that PC =QC. Also prove that Z MPC = Z CQM. C 8. In this figure, if AC = BC, AP = BQ, and PM=QM, prove that CM is ± to PQ. What angles must be proved to be right angles ? J-^ -^ Q~^B 9. In this figure P, Q, and R are mid-points of the sides of the equilateral triangle ABC. Prove that PQR is ^ an equilateral triangle. ^^ x g Prove that /kAPR, BQP, and CRQ are congruent by using two propositions already proved. a 34 BOOK I. PLANE GEOMETEY Proposition V. Theorem 76. If hvo angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle is isosceles. c c' A' B' Given the triangle ABC, with the angle A equal to the angle B. To prove that AC = BC. Proof. Suppose the second triangle A'B'C' to be an exact reproduction of the given triangle ABC. Turn the triangle A'B'C' over and place it upon ABC so that B' shall fall on A and A' shall fall on B. Post. 5 Then B'A' will coincide with AB. Post. 1 Since ZA' = ZB', Given and ZA=ZA', Hyp. .'.ZA=ZB'. Ax. 8 .'.^'C will lie along .1(7. Similarly A'C will lie along BC. Therefore C will fall on both AC and BC, and hence at their intersection. . . .'.B'C' = AC. But B'C' was made equal to BC, .'. AC = BC, by Ax. 8. q.e.d. 77. Corollary. An equiangular triangle is equilateral. TRIANGLES 35 78. Kinds of Proof. In the five propositions thus far proved in the text two different kinds of proof have been seen : (1) Synthetic. In Prop. I we put together some known truths in order to obtain a new truth. Such a method of proof is known as the synthetic method, and is the most common of all that are used in geometry. In this method we endeavor simply to find what propositions have already been proved that will lead to the proof of the proposition that is before us. This method was used in all the exercises on pages 28, 31, and 33. (2) By superposition. In Props. II and III we placed one figure on another and then, by synthetic reasoning, showed them to be identically equal. Such proof is known as a proof by super- position. Superposition means " placing on," and one figure is said to be superposed on the other. In Prop. V a special kind of proof by superposition was employed, in which we superpose a figure on its exact dupli- cate. This special method is rarely used, but in this proposi- tion it materially simplifies the proof. 79. Converse Propositions. If two propositions are so related that what is given in each is what is to be proved in the other, each proposition is called the converse of the other. E.g. in Prop. IV we have given AC = BC, to prove that ZA = ZB. In Prop. V we have given ZA = ZB, to prove that AC = BC. Hence Prop. V is tlie converse of Prop. IV, and Prop. IV is the con- verse of Prop. V. Not all converses are true, and hence we have to prove any given converse. E.g. tlie converse of the statement ''Two right angles are two equal angles" is "Two equal angles are two right angles," and this statement is evidently false. 36 BOOK I. pla:^^e geometry Proposition VI. Theorem 80. Tioo triangles are congruent if the three sides of the one are equal respectively to the three sides of the other. Given the triangles ABC and A'B'C\ with AB equal to A^B^ AC equal to A'C, and BC equal to B'C To prove that A ABC is congruent to AA'B'C. Proof. Let AB and A'B' be the greatest of the sides of the A. Place A A'B'C next to A ABC so that A' shall fall on A, the side A'B' shall fall along AB, and the vertex C' shall be opposite the vertex C. Post. 5 Then B' will fall on B. {For A'B'' is given equal to AB.) Draw CC. Since AC = AC', Given .•.ZACC' = ZCCA. §74 Since BC=BC', Given .'.ZC'CB = ZBC'C. §74 .-. Z .1 CC"+ Z C'CB = Z CCA + Z BC'C. Ax. 1 Hence ZylC5 = Z5C"^. Ax. 11 {For Z A CB is made up of ZACC and Z C'CB, and A BC'A is made up of Z CCA and Z BC'C.) .-. A ABC is congruent to A ABC'. § 68 .*. A.4J5C is congruent to A.l'^'C', by Ax. 9. q.e.d. TRIANGLES 37 EXERCISE 8 1. Prove that a line from the vertex to the mid-point of the base of an isosceles triangle cuts the triangle into two congruent triangles. 2. Three iron rods are hinged at the ex- tremities, as shown in this figure. Is the figure rigid ? Why ? 3. Four iron rods are hinged, as shown in this figure. Is the figure rigid ? If not, where would you put in the fifth rod to make it rigid ? Prove that this would accomplish the result. 4. If two isosceles triangles are constructed on opposite sides of the same base, prove by Prop. VI and § 58 that the line through the vertices bisects the vertical angles. 5. In this figure AB = AD and CB = CD. ^' Prove that AC bisects Z BAD and Z DCB. 6. In § 31, Ex. 8, it was shown how to bisect an angle, this being the figure used. Draw PX and PY, and prove by Prop. VI that PO bi- sects ZAOB. 7. In a triangle ABC it is known that A C = BC. It A A and Z B are both bisected by lines meet- ing at P, prove that AABP is isosceles. 8. In this figure it is known that Am = A n. Prove that ^C = ^C. 9. From the vertices A and B of an equilateral triangle lines are drawn to the mid-points of the opposite c sides. Prove that these two lines are equal. In A ABQ and BAP show that the conditions of congruence as stated in Prop. II are fulfilled. 38 BOOK L PLANE GEOMETRY Proposition VII. Theorem 81. The sum of tivo lines from a given point to the extremities of a given line is greater than the sum of tivo other lines similarly drawn, hut included hy them. B Given CA and C5, two lines drawn from the point C to the extremities of the line AB^ and PA and PB two lines similarly drawn, but included by CA and CB. To prove that CA-\-CB>PA-}- PB. Proof. Produce AP to meet the line CB at Q. Post. 2 Then CA -\- CQ>PA ^ PQ. Post. 3 {A straight line is the shortest path between two points.) Likewise .BQ-\-PQ>PB. Post. 3 Add these inequalities, and we have CAJrCQ + BQ + PQ>PA+PQ^PB. Ax. 7 {If unequals are added to unequals in the same order, the sums are unequal in the same order.) Substituting for CQ + BQ its equal CB, we have CA-\-CB-i- PQ >PA +PQ + PB. Ax. 9 {A quantity may he substituted for its equal in an equation or in an inequality.) Taking PQ from each side of the inequality, we have CA^CB>PA-\- PB, by Ax. 6. q. e. d. TEIANGLES 39 Proposition VIII. Theorem 82. Only one perpendicular can he drawn to a given line from a given external point. Given a line XF, P an external point, PO a perpendicular to XY from P, and PZ any other line from P to XY. To prove that PZ is not ± to XY. Proof. Produce PO to P', making OP' equal to PO. Post. 2 Draw P'Z. Post. 1 By construction POP' is a straight line. .'. PZP' is not a straight line. Post. 1 Hence Z P'ZP is not a straight angle. § 33 Since A POZ and ZOP' are rt. A, § 27 .'.ZP0Z = ZZ0P'. §56 Furthermore PO = OP', Hyp. and OZ — OZ. Iden. .-. A OPZ is congruent to AOP'Z, § 68 ( Two A are congruent if two sides and the included Z of the one are equal respectively to two sides and the included Z of the other.) and Z OZP = Z P'ZO. § 67 .*. Z OZP, the half of Z P'ZP, is not a right angle. § 34 .-. PZ is not ± to Xr, by § 27. Q.e.d. 40 BOOK I. PLANE GEOMETEY Proposition IX. Theorem 83. Tioo lines draion from a point in a perpendicu- lar to a given line, cutting off on the given line equal segments from the foot of the perpendicular, are equal and make equal angles ivith the perpendicular. A o B ^ Given PO perpendicular to XF, and PA and PB two lines cutting off on XY equal segments OA and OB from 0. To prove that PA = PB, and ZAPO = ZOPB. Proof. In the A A OP and BOP, Z POA and Z BOP are rt. A. § 27 {For PO is given ± to XY.) .'.ZP0A=ZB0P. §56 {All right A are equal.) Also OA = OB, Given and PO = PO. Men. {That is, PO is common to the two ^.) .-.A A OP is congruent to A BOP. § 68 ( Two A are congruent if two sides and the included Z of the one are equal respectively to two sides and the included Z of the other.) .-. PA = PB, and AAPO = Z OPB. § 67 {Corresponding parts of congruent figures are equal.) Q.E.D. TRIANGLES Proposition X. Theorem 41 84. Of two lines draion from a point in a perpen- dicular to a given line, cutting off on the given line unequal segments from the foot of the perpendicidar, the more remote is the greater. Given PO perpendicular to XY, PA and PC two lines drawn from P to XY, and OA greater than OC. To prove that PA>PC. Proof. Take OB equal to OC, and draw PB. Then PB = PC. § 83 Produce PO to P', making OP' = PO, and draw P'A and P'B. Then PA = P'A and PB = P'B. § 83 But PA + P'A >PB + P'B. § 81 . • . 2 PA >2PB and PA > PB. Axs. 9 and 6 .'.PA >PC, by Ax. 9. q.e.d. 85. Corollary. Onl^ two equal obliques can be drawn from a given point to a given line, and these cut off equal segments from the foot of the perpendicular. Of two unequal lines from a point to a line, the greater cuts off the greater segment from the foot of the perpendicular. For PB = PC, but PB cannot equal PA (§ 84). The segments OB and OC are equal, for otherwise PB could not equal PC 42 BOOK I. PLAKE GEOMETRY Proposition XI. Theorem 86. The perpendicular is the shortest line that can he drawn to a given line from a given external point. Given a line XF, P an external point, PO the peri)endicular, and PZ any other line drawn from P to XY. To prove that PO < PZ. Proof. Produce PO to P', making OP' = PO) and draw P'Z. Then PZ = P'Z. § 83 {Two lines drawn from a point in a 1. to a given line, cutting off on the given line equal segments from the foot of the ±, are equal.) .•.PZ + P'Z = 2 PZ. Ax. 1 Furthermore PO + P'O = 2 PO. Ax. 1 But PO + P'O <PZ-\- P'Z. Post. 3 .'.2PO<2PZ. Ax. 9 .'. PO<PZ, by Ax. 6. q.e.d. 87. Hypotenuse. The side opposite the right angle in a right triangle is called the hypotenuse. The other two sides of a right triangle are usually called the sides. 88. Distance. The length of the straight line from one point to another is called the distance between the points. The length of the perpendicular from an external point to a line is called the distance from the point to the line. TRIANGLES 43 Proposition XII. Theorem 89. Tico right triangles are congruent if the hypote- nuse and a side of the one are equal respectively to the hypotenuse and a side of the other. A / Given the right triangles ABC and A'5'C', with the h3rpotenuse AC equal to the hypotenuse A^C\ and with BC equal to 5'C'. To prove that A ABC is congruent to AA'B'C. Proof. Place A ABC next to A.4'^'C', so that BC shall fall along B'C, B shall fall on B'j and A and A ' shall fall on opposite sides of 5'C'. Then and Since C will fall on C, {For BC is given equal to B'C\) BA will fall along A 'B' produced. , {For A CBA + Z A'B'C'= a st. Z.) AC' = A'C\ Post. 5 §34 .•.AB' = A'B'. §85 .-. A ABC is congruent to A A^B^C. § 80 ( Two A are congruent if the three sides of the one are equal respectively to the three sides of the other.) Q. E. D. 90. Corollary. Two right triangles are congruent if any two sides of the one are equal respectively to the corresponding two sides of the other. 44 BOOK I. PLANE GEOMETRY EXERCISE 9 1. ABCD is a square and Mis the mid-point of AB. With ikf as a center an arc is drawn, cutting 5C at P and AD Sit Q. Prove that A MBP is congruent to A MA Q, and d c write the general statement of this theorem without using letters as is done here. This would read, "If an arc is drawn, with the mid- point of one side of a square as a center, cutting the ^ ^ sides perpendicular to that side, then the triangles cut off by," etc. 2. Draw a figure similar to that of Ex. 1, but take a radius such that the arc cuts BC produced at a point above C, and AD above D. Then prove that A MBP is congruent to A MA Q. 3. Prove that if from the point P the perpendiculars PM, PN to the sides of an angle A OB are equal, the point P lies on the bisector of the angle A OB. JVx Write the general statement of this theorem without using letters as is done here. ^ ^ 4. Prove that if the perpendiculars from the mid-point M of the base AB oi a, triangle ^i^C to the sides of the triangle are equal, then AA=A B. What then follows as to the sides A C and BC ? Write the gen- eral statement of this theorem without referring to a special figure. 5. Prove that if the perpendiculars from the extremities of the base of a triangle to the other two sides are equal, the triangle is isosceles. 6. Suppose 0Y1.0X. With as a center an arc is drawn cutting OX at A and OF at 5. Then with A as a center an arc is drawn cutting OF at P, and with 5 as a center and the same radius an arc is drawn cutting OX at Q. Prove that OP = OQ. What triangles are congruent by Prop. XII ? TRIANGLES 45 Proposition XIII. Theorem 91. Tico right triaiigles are congruent if the hypotenuse and an adjacent angle of the one are equal respectively to the hyjjotenuse and an adjacent angle of the other, c Given the right triangles ABC, A'B'C\ with the hypotenuse AC equal to the hypotenuse A'C, and with angle A equal to angle A'. To prove that A ABC is congruent to AA'B'C. Proof. Place A ABC upon A A'B'C so that A shall fall upon A' SiudAC shall fall along A 'C. Post. 5 Then C will fall onC, {For AC is given equal to A'C\) and AB will lie along A'B'. {For ZA is given equal to Z A'.) Then because C falls on C", and AB and B^ are rt. A, Given {Since the A are given as ri. A.) .-. CB will coincide with C'B'. § 82 {Only one perpendicular can be drawn to a given line from a given external point.) .'. A ABC is congruent to A A'B'C^. § 66 {If two figures can he made to coincide in all their parts, they are said to he congruent.) Q.E.D. 46 BOOK I. PLANE GEOMETRY Proposition XIV. Theorem 92. TiDO lines in the same plane perpendicular to the same line cannot meet hoiveve?' far they are produced. X Given the lines AB and CD perpendicular to XY at A and C respectively. To prove that AB and CD cannot meet however far they are produced. Proof, li AB and CD can meet if sufficiently produced, we shall have two perpendicular lines from their point of meeting to the same line. But this is impossible. § 82 .'. AB and CD cannot meet. q.e.d. 93. Parallel Lines. Lines that lie in the same plane and cannot meet however far produced are called parallel lines. 94. Postulate of Parallels. Through a given point only one line can he drawn parallel to a given line. As always in such cases the word line means straight line. 95. Corollary 1. Two lines iii the same plane perpen- dicular to the same line are parallel. 96. Corollary 2. Two lines in the same plane parallel to a third line are parallel to each other. For if they could meet, we should have two lines through a point parallel to a line. Why is this impossible ? PARALLEL LINES 47 Proposition XV. Theorem 97. If a line is j^erpendicular to one of tivo parallel lines, it is perpendicular to the other also. o P ^"^ Given AB and Ci>, two parallel lines, with XY perpendicular to AB and cutting CD at P. To prove that XY is ± to CD. Proof. Suppose MN drawn through P _L to XY. Then MN is II to AB. § 95 But CD is 11 to AB. Given .*. CD and MN must coincide. § 94 But XF is _L to MN. Hyp. .'. AT is X to CD. Q.B.D. 98. Transversal. A line that cuts two or more lines is called a transversal of those lines. 99 . Angles made by a Transversal. If XY cuts AB and CD, the angles a, d, g, f are called interior angles ; h, c, h, e are called exterior angles. The angles d and /, and a and g, are called alternate-interior angles ; c the angles b and h, and c and e, are called alternate-exterior angles. The angles b and/, c and g, e and a, h and d, are called exterior- interior angles. 48 BOOK I. PLAN^E GEOMETRY Proposition XVI. Theorem 100. If tivo parallel lines are cut hy a transversal, the alternate-interior angles are equal. X Given AB and CD, two parallel lines cut by the transversal XY in the points P and Q respectively. To prove that ZAFQ = ZDQP. Proof. Through 0, the mid-point of PQ, suppose MN drawn _L to CD. Then MN is likewise ±to AB. § 97 {A line ± to one of two \\s is 1. to the other.) Now A PMO and QNO are rt. A. § 63 {Since A OMP and ONQ are rt. A.) But Z POM = Z QON, § 60 (If two lines intersect, the vertical A are equal.) and OP=OQ. Hyp. {For was taken as the mid-point of PQ.) .'.A PMO is congruent to A QNO. § 91 ( Two right A are congruent if the hypotenuse and an adjacent Z of the one are equal respectively to the hypotenuse and an adjacent Z of the other.) .'.ZAPQ = ZDQP. §67 {Corresponding parts of congruent figures are equal.) Q. E. D. PAKALLEL LINES 49 Proposition XVII. Theorem 101. When two lines in the same plane are cut hy a transversal, if the alternate-interior angles are equal, the two lines are parallel. Y Given the lines AB and CD cut by the transversal XY in the points P and Q respectively, so as to make the angles APQ and DQP equal. To prove that AB is II to CD, Proof. Since we do not know that AB is II to CD, let us suppose MN drawn through P II to CD. We shall then prove that AB coincides with MN. Now Z MPQ = Z DQP. § 100 {If two II lines are cut by a transversal, the alt.-int. A are equal.) But Z APQ =:^^ DQP. Given .'.ZAPQ=:Z.MPQ. Ax. 8 {Quantities that are equal to the same quantity are equal to each other.) .'. AB and MN must coincide. § 23 (De/. of equal angles.) But MN is II to CD. Hyp. {For MN was drawn II to CD.) .'. AB, which coincides with MN, is II to CD. q.e.d. This proposition is the converse of Prop. XVI, as defined in § 79. 50 BOOK I. PLANE GEOMETEY Proposition XVIII. Theorem 102. If tivo parallel lines are cut hy a transversal, the exterior-interior angles are equal. Y Given AB and CD, two parallel lines, cut by the transversal XY in the points P and Q respectively. To prove that Z BPX =ADQX. Proof. Z BPX = Z.APQ. § 60 ZAPQ = ZDQX. §100 .'. ZBPX = ZDQX, by Ax. 8. q.e.d. 103. Corollary 1. When two lines are cut hy a transversal, if the exterior-interior angles are equal, the lines are parallel. The proofs of §§ 103 and 105 are similar to that of § 101. 104. Corollary 2. If two parallel lines are cut hy a trans- versal, the two interior angles on the same side of the trans- versal are supplementary. 105. Corollary 3. When two lines are cut hy a transversal, if two interior angles on the same side of the transversal are supplementary, the lines are parallel. 106. Corollary 4. If two parallel liries are cut hy a trans- versal, the alternate-exterior angles are equal. TEIANGLES 51 Proposition XIX. Theorem 107. The smn of the three angles of a triangle is equal to two right angles. c A Given the triangle ABC. To prove that AA-\-A B-\-ZC=2 H. A. Proof. Suppose BY drawn II to AC, and produce AB to X Then Z XB Y-\-Z.YBC + Z. CBA = 2 rt. A. § 34 {For a St. Z equals 2 rt. A.) But AA=AXBY, §102 and AC = ZYBC. §100 .'.ZA+AB-\-AC = 2Yt.A, by Ax. 9. -q.e.d. 108. Corollary 1. Tf two triangles have two angles of the one equal to two angles of the other, the third angles are equal. 109. Corollary 2. In a triangle there can he hut one right angle or one obtuse angle. 110. Exterior Angle. The angle included by one side of a figure and an adjacent side produced is called an exterior angle. In the above figure A XBC is an exterior angle, and A A and C are called the opposite interior angles. 111. Corollary 3. An exterior angle of a triangle is equal to the sum of the two opposite interior angles, and is therefore greater than either of them. 62 BOOK I. PLANE GEOMETRY EXERCISE 10 1. Show that if we place a draftsman's triangle against a ruler and draw A C, and move the triangle along as shown in the figure and draw A'C, then ^C is II to A'C. [ 2. In the next figure x = 60°. How many de- grees in each of the other seven angles ? 3. In the next figure representing two pairs of parallel lines certain angles are equal. State these equalities in this form : a = c = g = e = = •■', and give the reason in each case. 4. In the figure of Ex. 3 state ten pairs of nonadjacent angles that are supplementary. Thus : a-\-h= 180° and cZ + e == 180°. 5. In the triangle ABC, AC =BC and DE is drawn parallel to AB. Prove that CD = CE. Write a general statement of the theorem. 6. In the next figure A B is parallel to CD, and /-APQ is half of Z.QPB. How many degrees in the various angles ? 7. If Z YQD = 135°, how many degrees in the various angles ? 8. Let ZDQP = x and Z.YQD = y. Then if y — x = 100°, find the value of x and y. 9. Let ZCQY = x and ZXPA=y. the value of x and y. 10. In the next figure x = 72° and x = ^y. It is required to know if the lines are parallel, and why. 11. In the figure of Ex. 10 suppose x = 73° and y — x = 32°. It is required to know if the lines are parallel, and why. Then if x = ^y, find TRIANGLES 53 The three angles of a triangle are x, g, and z. Find the value of 0, given the values of x and y as follows : 12. X = 10°, y = 30°. 11. x = 37°, y = 48°. 13. a; = 20°, 2/ = 20°. 18. ic = 63°, y = 29°. 14. X = 75°, y = 50°. 19. a; =75° 29', y = 68° 41'. 15. X = 38°, y = 76°. 20. x = 82° 33', y = 75° 48'. 16. X = 49°, y = 92°. 21. ic = 69° 58', y = 82° 49'. 22. In a certain right triangle one angle is 37°. What is the size of the other acute angle ? 23. In a certain right triangle one angle is 36° 41'. What is the size of the other acute angle ? 24. In a certain right triangle one angle is 29° 48' 56". What is the size of the other acute angle ? 25. In a certain right triangle one acute angle is two thirds of the other. How many degrees are there in each ? 26. In a certain right triangle one acute angle is twice as large as the other. How many degrees are there in each ? 27. In a certain right triangle the acute angles are 2 x and 5 X. Find the value of x and the size of each angle. 28. In a certain triangle one angle is twice as large as another and three times as large as the third. How many degrees are there in each ? 29. In a certain isosceles triangle one angle is twice another angle. How many degrees in each of the three angles ? 30. In this figure what single angle equals a -\- G? To the sum of what angles is q equal ? also r ? From these relations find the number y^ ^ of degrees in ^9 + 5' + r. 31. Prove Prop. XIX by first drawing a parallel to AB through C, instead of drawing BY. 54 BOOK L PLANE GEOMETRY Proposition XX. Theorem 112. The sum of any two sides of a triangle is greater than the third side, and the difference hetiveen any tivo sides is less than the third side. A B Given the triangle ABC^ with AB the greatest side. To prove that BC + CA >AB, and AB -BC< CA. Proof. BC + CA >AB. Post. 3 {A straight line is the shortest path between two points.) Since BC-]-CA>AB, .■.CA>AB-BC; Ax. 6 or, AB — BC<CA. q.e.d. EXERCISE 11 State in what cases it is possible to form triangles with rods of the following lengths, and give the reason : 1. 2 in., 3 in., 4 in. 4. 7 in., 10 in., 20 in. 2. 3 in., 4 in., 7 in. 5. 8 in., 9^ in., 18 in. 3. 6 in., 7 in., 9 in. 6. 9f in., lOi in., 12i in. 7. In this figure prove that AB -\-BC >AD -]-DC. Why is DB+BODC? What is the result of adding AD to these unequals ? 8. How many degrees are there in each angle of an equiangular triangle ? Prove it. ^ ^ ^ TRIANGLES 55 Proposition XXI. Theorem 113. If two sides of a triangle are unequal , the angles opjjosite these sides are unequal, and the angle opposite the greater side is the greater. A B Given the triangle ABC^ with BC greater than CA. To prove that ZBAOZB. Proof. On CB suppose CX taken equal to CA. Draw AX. Post. 1 Then A AXC is isosceles. § 62 Then Z CXA = Z XA C. § 74 (In an isosceles A the A opposite the equal sides are equal.) But Z CXA >ZB. § 111 {An exterior Z of a A is greater than either opposite interior Z.) Also ZBAOZ XA C. Ax. 11 {For ZXAC isa part of Z BA C.) Substituting in this inequality for Z XAC its equal, Z CXA, we have the inequality Z BA C>Z CXA . Ax. 9 Since Z^^OZCA'.l, and Z.CXA>ZB, .'.ZBAOZB. Ax. 10 {If the first of three quantities is greater than the second, and the second is greater than the third, then the first is greater than the third.) Q. E. D. 56 BOOK T. PLANE GEOMETRY Proposition XXII. Theorem 114. If two angles of a triangle are unequal, the sides opposite these angles are unequal, and the side opposite the greater angle is the greater. c Given the triangle ABC^ with the angle A greater than the angle B. To prove that BOCA. Proof. Now BC is either equal to CA, or less than CA, or greater than CA. But ii BC were equal to CA, then the Z A would be equal to the Z.B. § 74 {For they would he A opposite equal sides.) And if CA were greater than BC, then the Z B would be greater than the Z.A. § 113 But if CA is not greater than BC, this is only another way of saying that BC is not less than CA. We have, therefore, two conclusions to be considered, ZA=ZB, and ZA<ZB. Both these conclusions are contrary to the given fact that the Z ^ is greater than the Z B. Since BC cannot be equal to CA or less than CA without violating the given condition, .'. BOCA. q.e.d. This proposition is the converse of Prop. XXL TRIANGLES Proposition XXIII. Theorem 57 115. If tivo triangles have two sides of the one equal respectively to two sides of the other, hut the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Y Y Given the triangles ABC and XYZ, with CA equal to ZX and BC equal to FZ, but with the angle C greater than the angle Z. To 2)rove that AB >XY. Proof. Place the A so that Z coincides with C and ZX falls along CA. Then .A' falls on A, since ZX is given equal to CA, and Z Y falls within Z A CB, since AC is given greater than Z Z. Suppose CP drawn to bisect the Z YCB, and draw YP. Then since CP = CP, CY = CB, Iden. Given d Zycp = Zpcb } Hyp. . '. A PYC is congruent to A PBC. §68 .•.PY=PB. §67 Now AP-{-PY>AY. .•.AP-^PB>AY. .\AB>AY. Post. 3 Ax. 9 Ax. 11 .\AB>XY, by Ax. 9. Q.E.D. 68 BOOK I. PLANE GEOMETRY Proposition XXIV. Theorem 116. If tivo triangles have two sides of the one equal respectively to tivo sides of the other ^ hut the third side of the first triangle greater than the third side of the second, then the angle opposite the third side of the first is greater than the angle opposite the third side of the second. z A B X Y Given the triangles ABC and XYZ^ with CA equal to ZX and BC equal to FZ, but with AB greater than XY. To prove that the Z. C is greater than the AZ. Proof. Kow the Z C is either equal to the Z Z, or less than the Z Z, or greater than the Z Z. But if the Z C were equal to the Z Z, then the A ^5C would be congruent to the A XFZ, § 68 {^or it would have two sides and the included Z of the one equal respectively to two sides and the included Z of the other.) and AB would be equal to XY. § 67 And if the Z C were less than the Z Z, then AB would be less than XY. § 115 Both these conclusions are contrary to the given fact that AB is greater than XY. .'. AOAZ. Q.E.D. This proposition is the converse of Prop. XXIII. QUADRILATERALS 59 117. Quadrilateral. A portion of a plane bounded by four straight lines is called a quadrllateraL 118. Kinds of Quadrilaterals. A quadrilateral may be a trapezoid, having two sides parallel ; 2^ parallelogram, having the opposite sides parallel. If the^nonparallel sides are equal, a trapezoid is called isosceles. A quadrilateral with no two sides parallel is called a trapezium. Trapezoid Parallelogram Tiai)eziuin 119. Kinds of Parallelograms. A parallelogram may be a rectangle, having its angles all right angles ; a rhombus, having its sides all equal. A parallelogram with all its angles oblique is called a rhomboid. Rectangle Rhombus Rhomboid 120. Base. The side upon which a hgure is supposed to rest is called the base. If a quadrilateral has a side parallel to the base, this is called the wpper base, the other being called the lower base. In an isosceles triangle the vertex formed by the equal sides is taken as the vertex of the triangle, and the side opposite this vertex is taken as the base of the triangle. 121. Altitude. The perpendicular distance between the bases of a parallelogram or trapezoid is called the altitude. The perpendicular distance from the vertex of a triangle to the base is called the altitude of the triangle. 122. Diagonal. The straight line joining two nonconsecutive vertices of any figure is called a diagonal. 60 BOOK I. PLANE GEOMETRY Proposition XXV. Theorem: 123. TiDO angles whose sides are parallel each to each are either equal or supplementary. z Given the angle AOB and the lines WY and XZ parallel to the sides and intersecting at P, the figure being lettered as shown. To prove that /.p = /LO, and that Z.p' is supplementary to ZO. Proof. Let OA meet XZ at M. Then in the figure ZO = Z 7/1, and Zp = Z m. § 102 {If two II lines are cut by a transversal, the ext.-int. A are equal.) .■.Zp = Z.O. Ax. 8 Also ZL])' is the supplement of Zp. § 42 .*. Zp' is supplementary to Z 0, by § 58. q.e.d. If the sides of two angles are parallel each to each, under what circumstances are the angles equal, and under what circumstances are they supplementary ? 124. Corollary. The opposite angles of a parallelogram are equal, and any two consecutive angles are supplementary. Draw the figure and explain how it is known that any angle is the supplement of its consecutive angle. If two opposite angles are supple- ments of the same angle, show that § 58 applies. QUADRILATERALS 61 Proposition XXVL Theorem 125. The opposite sides of a i^arallelogram are equal. A Given the parallelogram ABCD. To prove that BC = AD, and AB = DC. Proof. Draw the diagoDal A C. In the A ABC and CDA, AC = AC, Iden. ZBAC == ZDCA, and ZACB = Z CA D. § 100 .-. A ABC is congruent to A CDA. § 72 .-. BC = AD, and AB = DC, by § 67. O-e.d. 126. Corollary 1. A diagonal divides a parallelogram into two congruent triangles. Upon what theorem does this depend ? 127. Corollary 2. Segments of parallel lines cut off hy parallel lines are equal. How does this follow from the proposition ? 128. Corollary 3. Two parallellines are everywhere equally distant from each other. A B \i AB and CD are parallel, what can be said of Js dropped from any points in AB to CD (§ 127) ? Hence what may be said of all points in AB with respect to their distance from CD ? 62 BOOK L PLANE GEOMETRY Proposition XXVII. Theorem 129. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. Given the quadrilateral ABCD^ having BC equal to AD^ and AB equal to DC. To prove that the quadrilateral ABCD is a parallelogram. Proof. Draw the diagonal A C. In the A ABC ^nd'CDA, BC = AD, Given AB = DC, Given and AC = AC. Iden. .-. A ABC is congruent to A CDA. § 80 ( Two A are congruent if the three sides of the one are equal respectively to the three sides of the other.) .-. ZBAC==ZDCA, and ZACB = ZCAD. §67 .-. AB is II to DC, and BC is II to AD. § 101 {When two lines in the same plane are cut by a transversal, if the alt. -int. A are equal, the two lines are II.) .-.the quadrilateral ABCD is a O, by § 118. Q.e.d. This proposition is the converse of Prop. XXVI. QUADEILATERALS 63 Proposition XXVIII. Theorem 130. If two sides of a quadrilateral are equal and parallel, then the other two sides are equal and par- allel, and the figure is a p)arallelogram. Given the quadrilateral ABCD^ having AB equal and parallel to DC. To prove that the quadrilateral ABCD is a parallelogram. Proof. Draw the diagonal A C. In the A ABC and CDA, AC = AC, Iden. AB = DC, Given and ABAC = A DC A . § 100 {If two II lines are cut by a transversal, the alt.-int. A are equal.) .-.A ABC is congruent to A CDA. § 68 {Two A are congruent if two sides and the included Z of the one are equal respectively to two sides and the included Z of the other.) .\BC = AD, and ZACB = Z CAD. § 67 .-. BC is II to AD. § 101 ( When two lines in the same plane are cut by a transversal, if the alt.-int. A are equal, the two lines are II.) But AB is II to DC. Given .-. the quadrilateral ABCD is a O, by § 118. q.e.d. 64 BOOK I. PLANE GEOMETRY Proposition XXIX. Theorem 131. The diagonals of a parallelogram bisect each other. G A B Given the parallelogram ABCD , with the diagonals AC and BD intersecting at 0. To prove that A0 = OC, and B0 = 01). Proof. If we can show that the A ABO is congruent to the AC DO, or that the ABCO is congruent to the ADAO, the proposition is evidently proved, since the corresponding sides of the congruent triangles will be equal. Now in the A ABO and CDO, AB = CD, §125 {The opposite sides of a O are equal.) Z.BAO = Z.DCO, and Z OBA = Z ODC. § 100 {If two parallel lines are cut by a transversal, the alternate-interior angles are equal.) .'. AABOi^ congruent to A CDO. § 72 {Two A are congruent if two A and the included side of the one are equal respectively to two A and the included side of the other.) .\AO = OC, and BO = OD. § 67 {Corresponding parts of congruent A are equal.) Q. E. D. QUADRILATERALS 65 Proposition XXX. Theorem 132. Two j^ciraUelogranis are congruent if tivo sides and the included angle of the one are equal respectively to tivo sides and the included angle of the other. A B A' B' Given the parallelograms ABCD and A'B'C^D', with AB equal to A'B', AD to A'Z)', and angle A to angle A'. To prove that the UJ are congruent. Proof. Place the EJABCD upon the CJA'B'C'D' so that AB shall fall upon and coincide with its equal, A'B'. Post. 5 Then AD will fall along A'D\ (For Z A is given equal to Z A'.) and D will fall on D'. {For AD is given equal to A^D^.) Now DC and D'C are both II to A 'B' and are dl'awn through D'. .'.DC will fall along D'C. ' § 94 (Through a given point only one line can be drawn W to a given line.) Also BC and B'C' are both II to A 'D' and are drawn through B'. - .'. BC will fall along B'C. § 94 .-. C will fall on C. § 55 .'. the two [U coincide and are congruent, by § 66. q.e.d. 133. Corollary. Two rectangles having equal bases and equal altitudes are congruent. How is this shown to be a special case under the above proposition ? What sides are equal, and what inchided angles are equal ? 66 BOOK I. PLANE GEOMETRY Proposition XXXI. Theorem 134. If three or more parallels intercept equal, seg- ments on one transversal, they intercept equal segme7its on every transversal. Given the parallels AB^ CD^ EF, GH, intercepting equal segments JBD, DF, FH on the transversal BH^ and intercepting the segments AC J CEy EG on another transversal. To prove that AC = CE = EG. Proof. Suppose AP, CQ, and ER drawn II to BH. A APC, CQE, ERG = A BDC, DFE, FHG respectively. § 102 But A BDC, DFE, FHG are equal. § 102 .-. A APC, CQE, ERG are equal. Ax. 8 AP, CQ, ER are parallel. § 96 Also A CAP, ECQ, GER are equal. § 102 Now AP = BD, CQ = DF, ER = FH. § 127 {Segments of parallels cut off by parallels are equal.) But BD = DF=FH. Given .'. AP = CQ = ER. Ax. 8 .'. A CPA, EQC, and GRE are congruent. § 72 .'.AC = CE = EG, by § 67. Q.e.d. QUADRILATERALS 67 135. Corollary 1. If a line is parallel to one side of a tri- angle and bisects another side, it bisects the third side also. Let BE be II to BC and bisect AB. Suppose a line is drawn through A II to BC. Then how do we know this line to be II to DE ? Since it is given that the three lis intercept equal segments on the transversal AB, what can we say of the intercepted seg- ments on AC ? What can we then say that DE does to AC? Write the proof of this corollary in full. 136. Corollary 2. The line ivhich joins the mid-points of two sides of a triangle is parallel to the third side, and is equal to half the third side. A line DE drawn through the mid-point of AB, II to BC, divides AC in what way (§ 135) ? Therefore the line joining the mid-points oi AB and AC coincides with this parallel and is II to BC. Also since JPF drawn II to AB bisects AC, how does it divide BC ? What does this prove as to the relation of BF, FC, and BC ? Since BFED is a O (§118), what do we know as to the equality of DE, BF, and i BC ? Write the proof of this corollaiy in full. 137. Corollary 3. Tlie line joining the mid-points of the nonparallel sides of a trapezoid is parallel to the bases and is equal to half the sum of the bases. F^-^ A B Draw the diagonal DB. In the A ABD join E, the mid-point of AD, to F, the mid- point of DB. Then, by § 136, what relations exist between EF and AB? In the A DBC join F to G, the mid-point of BC. Then what relations exist between FG and DC? Since this relation exists, what relation exists between AB and FG ? But only one line can be drawn through F W to AB {% 94). Therefore FG is the prolongation of EF. Hence EFG is parallel to AB and CD, and equal to ^{AB-k- DC). Write the proof of this corollary in full. 68 BOOK I. PLAXE GEOMETRY 138. Polygon. A portion of a plane bounded by a broken line is called ?^ polygon. The terms sides, perimeter, angles, vertices, and diagonals are employed in the usual sense in connection with polygons in general. 139. Polygons classified as to Sides. A polygon is a triangle, if it has three sides ; a quadrilateral, if it has four sides ; a pentagon, if it has five sides ; a hexagon, if it has six sides. These names are sufficient for most cases. The next few names in order are heptagon, octagon, nonagon, decagon, undecagon, dodecagon. A polygon is equilateral, if all of its sides are equal. 140. Polygons classified as to Angles. A polygon is eqtdangular, if all of its angles are equal ; convex, if each of its angles is less than a straight angle ; concave, if it has an angle greater than a straight angle. Equilateral Equiangular Hexagon Convex Concave An angle of a polygon greater than a straight angle is called a reentrant angle. When the term polygon is used, a convex polygon is understood. 141. Regular Polygon. A polygon that is both equiangular and equilateral is called a regular polygon. 142. Relation of Two Polygons. Two polygons are mutually equiangular, if the angles of the one are equal to the angles of the other respectively, taken in the same order ; TRutually equilateral, if the sides of the one are equal to the sides of the other respectively, taken in the same order ; congruent, if mutually equiangular and mutually equilateral, since they then can be made to coincide. POLYGONS 69 Proposition XXXII. Theorem 143. The sum of the interior angles of a polygon is equal to two right angles, taken as r)iamj times less two as the figure has sides. Given the polygon ABCDEFy having n sides. To ])rove that the sum of the interior A =(n— 2^2 rt, A. Proof. From A draw the diagonals AC, AD, AE. The sum of the A of the A is equal to the sum of the A of the polygon. Ax. 11 Now there are (n — 2) A. {For there is one A for each side except the two sides adjacent to A.) The sum of the A of each A = 2 rt. ^. § 107 .*. the sum of the A of the (n — 2) A, that is, the sum of the A of the polygon, is equal to (n — 2)2 rt. A, by Ax. 3. q.e.d. 144. Corollary 1. The sum of the angles of a quadrilateral equals four right angles ; and if the angles are all equal, each is a right angle. 145. Corollary 2. Uach angle of a regular polygon of n sides is equal to — '- right angles. n 70 BOOK I. PLANE GEOMETRY EXERCISE 12 1. What is the sum of the angles of («) a pentagon ? (h) a hexagon ? (c) a heptagon ? (d) an octagon ? (6) a decagon ? (/) a dodecagon? (^) a polygon of 24 sides? 2. What is the size of each angle of (<() a regular pentagon ? (^) a regular hexagon ? (c) a regular octagon ? (c?) a regular decagon ? (e) a regular polygon of 32 sides ? 3. How many sides has a regular polygon, each angle of which is 1| right angles ? 4. How many sides has a regular polygon, each angle of which is If right angles ? 5. How many sides has a regular polygon, each angle of which is 108°? 6. How many sides has a regular polygon, each angle of which is 140°? 7. How many sides has a regular polygon, each angle of which is 156°? 8. Four of the angles of a pentagon are 120°, 80°, 90°, and 100° respectively. Find the fifth angle. 9. Five of the angles of a hexagon are 100°, 120°, 130°, 150°, and 90° respectively. Find the sixth angle. 10. The angles of a quadrilateral are x, 2x, 2x, and 3x. How many degrees are there in each ? 11. The angles of a quadrilateral are so related that the sec- ond is twice the first, the third three times the first, and the fourth four times the first. How many degrees in each ? 12. The angles of a hexagon are x, 3x, 3x, 2x, 2x, and x. How many degrees are there in each ? 13. The sum of two angles of a triangle is 100° and their difference is 40°. How many degrees are there in each of the three angles of the triangle ? POLYGONS 71 Proposttton XXXTir. Theorem 146. Tlie sum of the exterior angles of a polijcjon, made by j^'^oducing each of its sides hi succession, is equal to four right angles. Given the polygon ABCDE, having its n sides produced in succession. To prove that the sum of the exterior A = 4:rt. A. Proof. Denote the interior A of the polygon by a, b, c, d, e, and the corresponding exterior A by a', b', c', d', e'. Then, considering each pair of adjacent angles, A a -j- Aa'= a st. A, and Ab-{-Ab'= ^st A. §43 ( The two adjacent A which one straight line makes with another are together equal to a straight Z.) In like manner, each pair of adj. zi = a st. A. But the polygon has n sides and n angles. Therefore the sum of the interior and exterior A§ is equal to n st. A, OT 2n rt. A. Ax. 3 But the sum of the interior A = (n-2)2 rt. A § 143 = 2n rt. Zs - 4 rt. A. .*. the sum of the exterior A = 4: rt. A, by Ax. 2. q.e.d. 72 BOOK I. PLANE GEOMETRY EXERCISE 13 1. An exterior angle of a triangle is 130° and one of the opposite interior angles is 52°. Find the number of degrees in each angle of the triangle. 2. Two consecutive angles of a rectangle are bisected by lines meeting at P. How many degrees in the angle P ? 3. Two angles of an equilateral triangle are bisected by lines meeting at P. How many degrees in the angle P ? 4. The two base angles of an isosceles triangle are bisected by lines meeting at P. The vertical angle of the triangle is 30°. How many degrees in the. angle P? 5. The vertical angle of an isosceles triangle is 40°. This and one of the base angles are bisected by lines meeting at P. How many degrees in the angle P ? 6. One exterior angle of a parallelogram is one eighth of the sum of the four exterior angles. How many degrees in each angle of the parallelogram ? 7. How many degrees in each exterior angle of a regular hexagon ? of a regular octagon ? 8. In a right triangle one acute angle is twice the other. How many degrees in each exterior angle of the triangle ? 9. Make out a table showing the number of degrees in each interior angle and each exterior angle of regular polygons of three, four, five, • • • , ten sides. 10. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. 11. In this parallelogram ABCD, AP = CR, and BQ = DS. Prove that PQRS is ^ also a parallelogram. A F b 12. If the mid-points of the sides of a parallelogram are connected in order, the resulting figure is also a parallelogram. LOCI OF POINTS 73 147. Locus. The path of a point that moves in accordance with certain given geometric conditions is called the locus of the point. Thus, considering only figures in a plane, a . point at a given distance from a given line of ^ indefinite length is evidently in one of tw^o lines parallel to the given line and at the given distance from it. Thus, if ^jB is the given line and d the given distance, the locus is evidently the pair of parallel lines XY and X'Y\ ^ ^ The locus of a point in a plane at a given distance r / \ from a given point O is evidently the circle described about < r_\ as a center with a radius r. \ ^ I The plural of locus (a Latin v^^ord meaning "place") is \^ ^/ loci (pronounced lo-si). We may think of the locus as the place of all points that satisfy cer- tain given geometric conditions, and speak of the locus of points. Both expressions, locus of a point and locus of points, are used in mathematics. EXERCISE 14 State without proof the following loci in a plane : 1. The locus of a point 2 in. from a fixed point O. 2. The locus of the tip of the minute hand of a watch. 3. The locus of the center of the hub of a carriage wheel moving straight ahead on a level road. 4. The locus of a point 1 in, from each of two parallel lines that are 2 in. apart. 5. The locus of a point on this page and 1 in. from the edge. 6. The locus of the point of a round lead pencil as it rolls along a desk. 7. The locus of the tips of a pair of shears as they open, provided the fulcrum (bolt or screw) remains always fixed in one position. 8. The locus of the center of a circle that rolls around another circle, always just touching it. 74 BOOK I. PLANE GEOMETRY 148. Proof of a Locus. To prove that a certain line or group of lines is the locus of a point that fulhlls a given condition, it is necessary and sufficient to prove two things : 1. That any point In the sujyposed locus satisfies the condition. 2. That any point outside the supposed locus does not satisfy the given condition. For example, if we wish to find the locus of a point equidistant from these intersecting lines AB^ CD^ it is not sufficient to prove that any point on the angle-bisector PQ is equidistant from AB and CD^ because this may be only part of the locus. It is necessary to prove that no point outside of PQ satisfies the condition. In fact, in this case there is another line in the locus, the bisector of the Z BOD, as will be shown in § 152. 149. Perpendicular Bisector. A line that bisects a given line and is perpendicular to it is called the per2Je7idicida7' bisector of the line. EXERCISE 15 Draw the following loci, giving no proofs : 1. The locus of a point \ in. below the base of a given triangle ABC. 2. The locus of a point ^ in. from a given line AB. 3. The locus of a point 1 in. from a given point O. 4. The locus of a point ^ in. outside the circle described about a given point O with a radius 1\ in. 5. The locus of a point \ in. within the circle described about a given point with a radius 1^ in. 6. The locus of a point ^ in. from the circle described about a given point with a radius 1^ in. 7. The locus of a point ^ in. from each of two given parallel lines that are 1 in. apart. LOCI OF POINTS 75 Proposition XXXIV. Theorem 150. The locus of a point equidistant from the extrem- ities of a given line is the perpendicular bisector of that line. ^ b['\ Given FO, the perpendicular bisector of the line AB. To prove that YO is the locus of a point equidistant from A and B. Proof. Let P be any point in YO, and C any point not in YO. Draw the lines PA, PB, CA, and CB. Since . AO = BO, Given and OP = OP, Iden. .-. rt. A yl OP is congruent to rt. A BOP. § 90 .\PA=PB. §67 Let CA cut the X at D, and draw DB. Then, as above, DA = DB. But. CB<CD-{-DB. - '— Post. 3 .-. CB<CD + DA. Ax. 9 .'. CB<CA. Ax. 11 .*. FO is the required locus, by § 148. q.e.d. 151. Corollary. Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line. 76 BOOK I. PLANE GEOMETRY Proposition XXXV. Theorem 152. The locus of a point equidistant from tivo given intersecting lines is a pair of lines hisecting the angles formed hy those lines. Given XX^ and YY^ intersecting at O, -4C the bisector of angle X'OF, and BD the bisector of angle YOX. To prove that the pair of lines AC and BD is the locus of a point equidistant from XX' and YY', Proof. Let P be any point on A C or BD, and Q any point not on AC or BD. Let PM and QR be X to XX', PN and QS to YY'. Since Z MOP = Z PON, Given and OP = OP, Iden. .'. rt. A OMP is congruent to rt. A ONP. § 91 .\PM=PN. §67 Let QS cut AO at P'. Draw PT J_ to XX', and draw QT. Then, as above, P'T= P'S. But P'r-{-P'Q>QT, Post. 3 and QT>QR. §86 .-. P'T-\-P'Q>QR. Ax. 10 Substituting, P'5 + ^'Q > QR, or Q^^ > Q/?. Ax. 9 .'. the pair of lines is the required locus, by § 148. q.e.d. METHODS OF PKOOF 77 153. The Synthetic Method of Proof. The method of proof in which known truths are put together in order to obtain a new truth is called the synthetic method. This is the method used in most of the theorems already given. The proposition usually suggests some known propositions already proved, and from these v^^e proceed to the proof required. The exercises on this page and on- pages 78 and 79 may be proved by the synthetic method. 154. Concurrent Lines. If two or more lines pass through the same point, they are called concurrent lines. 155. Median. A line from any vertex of a triangle to the mid-point of the opposite side is called a median of the triangle. EXERCISE 16 1. If two triangles have two sides of the one equal respec- tively to two sides of the other, and the angles opposite two equal sides equal, the angles opposite the other two equal sides are equal or supplementary, and if equal the triangles are congruent. Let ^C = A'C\ BC = B'C\ and /.B = /.B\ Place AA'B'C on A ABC so that B'C shall coincide with BC, and ZA^ and ZA shall be on the same side of BC. Since ZB'= ZB, B'A' will fall along what line ? Then A' will fall at A or at some other point in BA, as B. If A' falls at A, what do we know about the congruency of the AA'B'C and ABC ? If A^ falls at D, what about the congruency of the A A'B'C and BBC ? Since CB = C'A' = CA, what about the relation of Z ^ to Z CBA ? Then what about the relation of the A CBA and BBC ? Then what about the relation of the A A and BBC ? Draw figures and show that the triangles are congruent : 1. If the given angles B and JB' are both right or both obtuse. 2. If the angles A and A^ are both acute, both right, or both obtuse* 3. If ^C and A'C are not less than BC and B'C respectively. 78 BOOK I. PLANE CxEOMETEY 2. The bisectors of the angles of a triangle are concurrent in a point equidistant from the sides of the triangle. The bisectors of two angles, as AD and BE, intersect as at 0. Why? Now show that is equidistant from AC and ^ AB, also from BC and AB, and hence from AC and BC. Therefore, where does lie with respect to the ,_g,, p bisector Ci^? ^^l^^ This point is called the incenter of the triangle. ^ p 3. The perpendicular bisectors of the sides of a triangle are concurrent in a point equidistant from the vertices. The ± bisectors of two sides, as QQ^ and RW, intersect as at 0. Why ? Now show that is equidistant from B and O, also from C and A, and hence from A and B. Therefore, where does lie with respect to the ± bisector PP' ? This point is called the circumcenter of the triangle. 4. The perpendiculars from the vertices of a triangle to the opposite sides are concurrent. Let the Js be J.Q, BR, and CP. Through A, B, C suppose B'C% C'A\ Bx -lA' & and ^'i^' drawn II to CB, AC, and BA respec- tively. Now show that C" J. ==50 = ^B'. In the same way, what are the mid-points of C'A' and A'B' ? How does this prove that A Q, BR, and CP J are the _L bisectors of the sides of the A A''B'C' ? Proceed as in Ex. 3. This point is called the orthocenter of the triangle: 5. The medians of a triangle are concurrent in a point two thirds of the distance from each vertex to the middle of the opposite side. Two medians, as ^ Q and CP, meet as at 0. If Y is the mid-point of A 0, and X of CO, show that YX and PQ are II to ^C and equal to ^ AC. Then show that AY=YO=OQ, and CX = XO = OP. Hence any median cuts off on any other median what part of the distance from the ver- tex to the mid-point of the opposite side ? This point is called the centroid of the triangle. METHODS OF PROOF 79 6. The bisectors of two vertical angles are in b the same straight line. ol^^^ c — ^^ — -^ 7. The bisector of one of two vertical angles ^/</ bisects the other. d 8. The bisectors of two supplementary adjacent angles are perpendicular to each other. 9. The bisectors of the two pairs of vertical angles formed by two intersecting lines are perpendicular to each other. 10. If the bisectors of two adjacent angles are JY .b perpendicular to each other, the adjacent angles \/^^ are supplementary. o 11. If an angle is bisected, and if a line is drawn ^ ^^^ ^ through the vertex perpendicular to the bisector, this line forms equal angles with the sides of the given angle. o ^ 12. The bisector of the vertical angle of an isosceles triangle bisects the base and is perpendicular to the base. 13. The perpendicular bisector of the base of an isosceles triangle passes through the vertex and bisects the e angle at the vertex. 14. If the perpendicular bisector of the base of a triangle passes through the vertex, the triangle is isosceles. a d b 15. Any point in the bisector of the vertical angle of an isos- celes triangle is equidistant from the extremities of the base. 16. If two isosceles triangles are on the same base, a line passing through their vertices is perpendicular to the base and bisects the base. 17. Two angles whose sides are perpendicular each to each are either equal or supplementary. Under what circumstances are the angles equal, and under what circumstances are they supplementary ? 80 BOOK I. PLANE GEOMETKY 156. The Analytic Method of Proof. The method of proof that asserts that a proposition under consideration is true if another proposition is true, and so on, step by step, until a known truth is reached, is called the analytic method. This is the method resorted to when we do not see how to start the ordinary synthetic proof. The exercises on this page and on pages 81 and 82 may be investigated by the analytic method. EXERCISE 17 1. The mid-point of the hypotenuse of a right triangle is equidistant from the three vertices. Given M the mid-point of AC, the hypotenuse of the rt. /\ABC. To prove that M is equidistant from A, B, and C. We may reason thus : M is equidistant from A, B, and C if AM= BM. Why is this the case ? AM= BM if the J. MN cuts A ABM into two congruent A. ]\] A ANM is congruent to A BNM ii AN= NB. But AN does equal NB (§ 135), because MN is II to CB, and AM = MC. Therefore the proposition is true. We may now, in writing our proof, begin with this last step and work backwards, as in the synthetic proofs already considered. 2. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter side. Given ZA = Z a, and ZC = Z2a, to prove that ^C is double BC. Let M be the mid-point of AC. Then ^C is double BC if AM=BC. Why ? Now if we draw MN II to CB, what can be said of the relation of AN and NB ? Why ? Then what may be said of A ANM and BNM? Why ? Then what may be said of AM &nd BM ? of Za and Zq? Therefore the proposition is true if BM = BC. But BM= BC ii Z2a = Zr, or if Z2 a = Za-{- Zq, or it Za = Zq. But Za = Zq because we have proved that AM = BM. Now reverse this reasoning and write the proof in the usual synthetic form. METHODS OF PROOF 81 3. A median of a triangle is less than half the sum of the two adjacent sides. Given CM a median of the A ABC. a To prove that CM<i{BC -\- CA). Now CM<i{BC -\-CA), if 2CM<BC + CA. "\ /J/ This suggests producing CM by its own length to P, \ / and drawing AP. P Then CP = 2 CM, and 2 CM<BC + CA if CP<BC -\-CA. But CP<^P + Cy1. Post. 3 .-. CP<BC-\- CA if BC = AP, and BC = AP if A JlfBC is congraent to A MAP. § 67 But A MBC is congraent to A MAP, § 68 for MB = MA, Given CM=MP, Hyp. and ^JBJfC = Z^3fP. \ §60 .■.CP<BC-{-CA. .'.CM<i{BC + CA). 4. The line which bisects two sides of a ti'iangle is parallel to the third side. ^ Given AD equal to DB, and AE equal to EC. To prove that DE is II to BC. Df Suppose a Une drawn from C II to BA, and suppose DE produced to meet it at G. DE is II to BC if BCGD is a O. BCGD is a O if CG = BD. CG = BD if each is equal to AD. Now BD = AD, and CG = ^D if A CGE is congraent to A ADE. But A CG^-E is congraent to A ADE, for EC = AE, Z. CEG = Z AED, and ZGCE = ZA. 82 BOOK I. PLANE GEOMETRY 5. Two isosceles triangles are congruent if a side and an angle of the one are equal respectively to the corresponding side and angle of the other. The A are congruent if what three corresponding parts are equal ? 6. The bisector of an exterior angle of an isosceles tri- angle, formed by producing one of the equal sides through the vertex, is parallel to the base. AE is II to BC if what angles are equal ? These angles are equal if Z CAD is twice what angle in the A ? ^ ^^ 7. If one of the equal sides of an isosceles triangle is pro- duced through the vertex by its own length, the line joining the end of the side produced to the nearer end of the base is perpendicular to the base. Z DBA is a rt. Z if it equals the sum of what A of A ABD ? ^ It equals this sum if Zp equals what angle and Z q equals what other angle ? 8. If the equal sides of an isosceles triangle are produced through the vertex so that the external segments are equal, the extremities of these segments are equidistant from the extremities of the base respectively. 9. If the line drawn from the vertex of a triangle to the mid-point of the base is equal to half the base, the angle at the vertex is a right angle. 10. If through any point in the bisector of an / ^^ ^ angle a line is drawn parallel to either side of the angle, the triangle thus formed is isosceles, o 11. Through any point C in the line AJ3 an intersecting line is drawn, and from any two points in this line equidistant from C perpendiculars are drawn to AB ov AB produced. Prove that these perpendiculars are equal. 12. The lines joining the mid-points of the sides of a triangle divide the triangle into four congruent ^"^ ^"^ triangles. METHODS OF PEOOF 83 157. The Indirect Method of Proof. The method of proof that assumes the proposition false and then shows that this assump- tion is absurd is called the indirect method or the reductio ad ahsurdum. This method forms a kind of last resort in the proof of a proposition, after the synthetic and analytic methods have failed. EXERCISE 18 1. Given ABC and ABD, two triangles on the same base AB, and on the same side of it, the vertex of each triangle being outside the other triangle. Prove that ii AC equals AD, then BC cannot equal BD. Assume that BC = BD and show that the result is absurd, since it would make D fall on C, which is contrary to the given conditions. 2. On the sides of the angle XOY two equal segments OA and OB are taken. On AB Si triangle APB is constructed with AP greater than J5P. Prove that OP cannot y bisect the angle XOY. ^ <r^'^^'^^ Assume that OP does bisect ZXOY. What is ^^^^J/"^ the result ? Is this result possible ? ^ 3. From M, the mid-point of a line AB, MC is drawn oblique to AB. Prove that CA cannot equal CB. c Assume that CA does equal CB. What is the result ? Is this result possible ? 4. If perpendiculars are drawn to the sides ^ m ^ of an acute angle from a point within the angle, they cannot inclose a right angle or an acute angle. Assume that they inclose a right angle and show that this leads to an absurdity. Similarly for an acute angle. 5. One of the equal angles of an isosceles triangle is five ninths of a right angle. Prove that the angle at the vertex cannot be a right angle. - Assume that it is a right angle. Is the result possible ? 84 BOOK I. PLANE GEOMETKY 158. General Suggestions for proving Theorems. The following general suggestions will often be helpful : 1. Draw the figures as accurately as possible. This is especially helpful at first. A proof is often rendered difficult simply because the figure is carelessly drawn. If one line is to be laid off equal to another, or if one angle is to be made equal to another, do this by the help of the compasses or by measuring with a ruler. 2. Draw as general figures as possible. If you wish to prove a proposition about a triangle, take a scalene tri- angle. If an equilateral triangle, for example, is taken, it may lead to believing something true for every kind of a triangle, when, in fact, it is true for only that particular kind. 3. After drawing the figure state very clearly exactly what is given and exactly tvhat is to be proved. Many of the difficulties of geometry come from failing to keep in mind exactly what is given and exactly what is to be proved. 4. Then proceed synthetically with the proof if you see how to begin. If you do not see how to begin, try the analytic method, stating clearly that you could prove this if you could prove that, and so on until you reach a known proposition. 5. If two lines are to be proved equal, try to prove them corre- sponding sides of congruent triangles, or sides of an isosceles triangle, or opposite sides of a parallelogram, or segments between parallels that cut equal segments from another transversal. 6. If two angles are to be proved equal, try to prove them alternate- interior or exterior-interior angles of parallel lines, or corresponding angles of congruent triangles, or base angles of an isosceles triangle, or opposite angles of a parallelogram. 7. If one angle is to be j^^oved greater than another, it is prob- ably an exterior angle of a triangle, or an angle opposite the greater side of a triangle. 8. If one line is to be proved greater than another, it is prob- ably opposite the greater angle of a triangle. EXERCISES 85 EXERCISE 19 Prove the following propositions referring to equal lines : 1. If the sides AB and AD oi ?, quad- rilateral ABCD are equal, and if the di- agonal AC bisects the angle at A, then EC is equal to DC. A B 2. A line is drawn terminated by two parallel lines. Through its mid-point any line is drawn terminated by the parallels. Prove that the second line is bisected by the first. 3. In a parallelogram ABCD the line BQ bisects AD, and DP bisects BC. Prove that Q, BQ and DP trisect AC. p X 4. On the base .45 of a triangle ABC any point P is taken. The lines AP, PB, BC, and CA are bisected by W, X, Y, and Z respec- tively. Prove that X F is equal to WZ. ^ w 5. In an isosceles triangle the medians drawn to the equal sides are equal. 6. In the square ABCD, CD is bisected by Q, and P and R are taken on AB so that AP equals BR. Prove that PQ equals RQ. c 7. In this figure AC = BC, and ^P = BQ = CR = CS. Prove that QR = PS. 8. From the vertex and the mid-points of the equal sides of an isosceles triangle lines are drawn perpendicular to the base. Prove that they divide the base into four equal parts. 9. In the quadrilateral ABCD it is known that AB is parallel to DC, and that angle C equals angle D. On CD two points are taken such that CP=DQ. Prove that AP = BQ. D Q PC c 86 BOOK I. PLANE GEOMETRY EXERCISE 20 Prove the following propositions referring to equal angles : 1. In this figure it is given that AC = BC, and that BQ and AR bisect the angles YBC and CAX respectively. Prove that AAPB is isosceles. 2. If through the vertices of an isosceles triangle lines are drawn parallel to the oppo- site sides, they form an isosceles triangle. 3. If the vertical angles of two isosceles triangles coincide, the bases either coincide or are parallel. 4. In which direction must the side of a triangle be produced so as to intersect the bisector of the opposite exterior angle ? Consider the cases, ZA<ZC,ZA = ZC,ZA>ZC. 5. The bisectors of the equal angles of an isosceles triangle form, together with the base, an isosceles triangle. 6. The bisectors of the base angles of an equilateral triangle form an angle equal to the exterior angle at the vertex of the triangle. 7. If the bisector of an exterior angle of a triangle is parallel to the opposite side, the tri- angle is isosceles. 8. A line drawn parallel to the base of an isosceles triangle makes equal angles with the sides or the sides produced. 9. A line drawn at right angles to AB, the base of an isosceles triangle ABC, cuts ^C at P and BC produced at Q. Prove that PCQ is an isosceles triangle. b d 10. In this figure, it AB = CD, and Z ^ = Z C, \ / then BD is parallel to AC. \ J, EXERCISES 87 EXERCISE 21 Prove the following propositions hy showing that two tri- angles are congruent : 1. A perpendicular to the bisector of an angle forms with the sides an isosceles triangle. 2. If two lines bisect each other at right angles, any point in either is equidistant from the extremities of the other, 3. From B a perpendicular is drawn to the bisector of the angle A of the triangle ABC, meeting it at X, and meeting ^C or AC produced at Y. Prove that BX = XY. 4. If through any point equally distant from two parallel lines two lines are drawn cutting the parallels, they intercept equal segments on these parallels. ^ 5. If from the point where the bisector of an angle of a triangle meets the opposite side, parallels are drawn to the other two sides, and terminated by the sides, these parallels are equal. 6. The diagonals of a square are perpendicular to each other and bisect the angles of the square. 7. If from a vertex of a square there are drawn line-seg- ments to the mid-points of the two sides not adjacent to the vertex, these line-segments are equal. 8. If either diagonal of a parallelogram bisects one of the angles, the sides of the parallelogram are Q all equal. / \^ 9. On the sides of any triangle ABC equi- / \^ lateral triangles BPC, CQA, ARB are con- /^.^--^/X structed. Prove that AP = BQ = CR. ^\~~^— / ^ P \ /'^ How can we prove that A ABP is congruent to \ / ARBC? Also that A ^fiC is congruent to A ^^Q? V/ Does this prove the proposition ? \B " " -'' 88 BOOK I. PLANE GEOMETRY EXERCISE 22 Prove the following propositions relating to the sum of the angles of a polygon : 1. An exterior angle of an acute triangle or of a right triangle cannot be acute. 2. If the sum of two angles of a triangle equals the third angle, the triangle is a right triangle. 3. If the line joining any vertex of a triangle to the mid- point of the opposite side divides the triangle into two isos- celes triangles, the original triangle is a right triangle. 4. If the vertical angles of two isosceles triangles are sup- plements one of the other, the base angles of the one are complements of those of the other. 5. From the extremities of the base AB oi Si triangle ABC perpendiculars to the other two sides are drawn, meeting at P. Prove that the angle P is ^ the supplement of the angle C. 6. If two sides of a quadrilateral are parallel, and the other two sides are equal but not parallel, the sums of the two pairs of opposite angles are equal. 7. The bisectors of two consecutive angles of a parallelogram are perpendicular to each other. 8. The exterior angles at B and C of any triangle ABC are bisected by lines meeting at P. Prove that the angle at P together with half the angle A equals a right angle. 9. The opposite angles of the quadrilateral formed by the bisectors of the interior angles of any quadrilateral are supplemental. 10. Show that Ex. 9 is true, if the bisectors of the exterior angles are taken. EXERCISES 89 EXERCISE 23 Prove the following propositions referring to greater lines or greater angles : 1. In the triangle ABC the angle A is bisected by a line meeting BC at D. Prove that BA is greater than BD, and CA greater than CD. 2. In the quadrilateral ABCD it is known that AD is the longest side and BC the shortest side. Prove that the angle B is greater than the angle D, and the angle C greater p than the angle A. ^ 3. A line is drawn from the vertex y4 of a square ABCD so as to cut CD and to meet BC produced in P. Prove that ^P is greater than DB. 4. If the angle between two adjacent sides of a parallelo- gram is increased, the length of the sides remaining unchanged, the diagonal from the vertex of this angle is diminished. c 5. Within a triangle ABC a point P is taken such that CP = CB. Prove that AB is always greater than A P. ^ A M B 6. In a quadrilateral ABCD it is known that AD equals BC and that the angle C is less than the angle D. Prove that the diagonal ^C is greater than the diagonal BD. 7. In the quadrilateral ABCD it is known that AD equals BC and that the angle D is greater than the angle C. Prove that the angle B is greater than the angle .4, c 8. In the triangle ABC the side AB is greater ^/ than A C. On A B and A C respectively BP is taken equal to CQ. Prove that Z>Q is greater than CP. ^ ^ 9. The sum of the distances of any point from the three vertices of a triangle is greater than half the sum of the sides. 90 BOOK I. PLANE GEOMETRY EXERCISE 24 Prove the following miscellaneous exercises : 1. The line joining the mid-points of the nonparallel sides of a trapezoid passes through the mid-points of jy the two diagonals. How is EF related to AB and BC ? Why ? Since EF bisects BC and AB, how does it divide AC aiidBB? Why? 2. The lines joining the mid-points of the consecutive sides of any quadrilateral form a parallelogram. How are PQ and SR related to ^C ? ^ p b 3. If the diagonals of a trapezoid are equal, the trapezoid is isosceles. c, d Draw CE and BF ± to AB. How isAABF related to A BCE ? Why ? Then how is Z FAB related to Z CBA ? Then how \sAABC related to A BAB? Why ? a E F B 4. If from the diagonal DB, of a square ABCD, BE is cut off equal to 5C, and EF is drawn perpendicular to BDj meeting DC 2X F, then DE is equal to jBi^ and also to FC. How many degrees in A EBF and BFE ? How is BE related to J^F? Why? Then how is rt. A BEF related to rt. A BCF ? Why ? 5. If the opposite sides of a hexagon are equal and parallel, the diagonals that join opposite vertices meet in a point. 6. If perpendiculars are drawn from the four vertices of a parallelogram to any line outside the parallelogram, the sum of the perpendiculars from one pair of opposite vertices equals the sum of those from the other pair. How are x ■\- y and to -|- 2- related to A: ? z y A k /y/ EXERCISES 91 EXERCISE 25 Examination Questions 1. The sum of the four sides of any quadrilateral is greater than the sum of the diagonals. 2. The lines joining the mid-points of the sides of a square, taken in order, form a square. 3. In a quadrilateral the angle between the bisectors of two consecutive angles is one half the sum of the other two angles. 4. If the opposite sides of a hexagon are equal, does it follow that they are parallel ? Give reasons for your answer. 5. In a triangle ABC the side BC is bisected at P and AB is bisected at Q. AP is produced to R so that AP= PR, and CQ is produced to S so that CQ = QS. Prove that S, B, and it are in a straight line. 6. If the diagonals of a parallelogram are equal, all of the angles of the parallelogram are equal. 7. In the triangle ABC, ZA = 60° and ZB>ZC. Which is the longest and which is the shortest side of the triangle ? Prove it. 8. How many sides has a polygon each of whose interior angles is equal to 175° ? 9. Given the quadrilateral ABCD, with AB equal to AD, and BC equal to CD. Prove that the diagonal A C bisects the angle DCB and is perpendicular to the diagonal BD. 10. In how many ways can two congruent triangles be put together to form a parallelogram ? Draw the diagrams. 11. The sides of a polygon of an odd number of sides are produced to meet, thus forming a star-shaped figure. What is the sum of the angles at the points of the star ? The propositions in Exercise 25 are taken from recent college entrance examination papers. 92 BOOK I. PLANE GEOMETRY EXERCISE 26 Review Questions 1. Define and illustrate rectilinear and curvilinear figures. 2. Upon what does the size of an angle depend ? 3. What is meant by the bisector of a magnitude ? Illus- trate when the magnitude is a line ; an angle. 4. Define perpendicular and state three facts relating to a perpendicular to a line. 5. Name and define the parts of a triangle and such special lines connected with a triangle as you have thus far studied. 6. Classify angles. 7. Classify triangles as to angles ; as to sides. 8. Define and illustrate complementary, supplementary, and conjugate angles. 9. What are the two classes of assumptions in geometry ? Give the list of each. 10. State all of the conditions of congruency of two triangles. 11. What is meant by the converse of a proposition ? 12. Are two triangles always congruent if three parts of the one are respectively equal to three parts of the other ? 13. State three tests for determining whether one line is parallel to another. 14. State the proposition relating to the sum of the angles of a triangle, and state a proposition that can be proved by its use. 15. State a proposition relating to two unequal angles of a triangle ; to two unequal sides of a triangle. 16. Must a triangle be equiangular if equilateral ? Must a triangle be equilateral if equiangular ? 17. Classify polygons as to sides ; as to angles. 18. Define locus and give three illustrations. BOOK II THE CIRCLE 159. Circle. A closed curve lying in a plane, and such that all of its points are equally distant from a fixed point in the plane, is called a circle. 160. Circle as a Locus. It follows that the locus of a point in a plane at a given distance from a fixed point is a circle. 161. Radius. A straight line from the center to the circle is called a radius. 162. Equal Radii. It follows that all radii of the same circle or of equal circles are equal, and that all circles of equal radii are equal. 163. Diameter. A straight line through the center, termi- nated at each end by the circle, is called a diameter. Since a diameter equals two radii, it follows that all diameters of the same circle or of equal circles are equal. 164. Arc. Any portion of a circle is called an arc. An arc that is half of a circle is called a semicircle. An arc less than a semicircle is called a minor arc, and an arc greater than a semicircle is called a major arc. The word arc taken alone is gen- erally understood to mean a minor arc. 165. Central Angle. If the vertex of an angle is at the center of a circle and the sides are radii of the circle, the angle is called a central angle. An angle is said to intercept any arc cut off by its sides, and the arc is said to subtend the angle. 94 BOOK II. PLANE GEOMETEY Proposition I. Theorem 166. Ill the same circle or in equal ciixles equal cen- tral angles intercept equal arcs ; and of tivo unequal central angles the greater intercepts the greater arc. Given two equal circles with centers and 0\ with angles AOB and A'O'B' equal, and with angle AOC greater than angle A'O'B^. To prove that 1. arc AB = are A'B'; 2. arc AC> arc A'B'. Proof. 1. Place the circle with center O on the circle with center 0' so that Z AOB shall coincide with its equal, AA'O'B'. In the case of the same circle, swing one angle about until it coincides with its equal angle. Then A falls on A', and B on B'. {Radii of equal circles are equal.) .'. arc AB coincides with arc A'B'. {Every point of each is equally distant from the center.) Proof. 2. Since Z^OC is greater than Z^'0'5', and ZAOB = Z.A 'O'B', therefore . Z^OC is greater than Z. 4 (95. Therefore OC lies outside ZAOB. .'. 3iTG AC>3i,TGAB. But ' arc ^5 = arc ^'^'. .*. arc ^ C> arc A 'B', by Ax. 9. q.e.d. Post. 5 §162 §159 Given Given Ax. 9 Ax. 11 CENTRAL ANGLES 95 Proposition II. Theorem 167. In the same circle or in equal circles equal arcs subtend equal central angles ; and of two unequal arcs the greater subtends the gi^eater central angle. Given two equal circles with centers and O ', with arcs AB and A^B^ equal, and with arc AC greater than arc -4 '5'. To prove that 1. AAOB = Z. A'O'B'; 2. ZAOOZ. A'O'B'. Proof. 1. Using the figure of Prop. I, place the circle with center O on the circle with center O' so that OA shall fall on its equal O'A', and the arc AB on its equal A'B'. Post. 5 Then OB coincides with O'B'. Post. 1 .■.ZAOB = Z A'O'B'. §23 Proof. 2. Since arc AC> arc A'B', it is greater than arc AB, the equal of arc A'B', and OB lies within the Z.AOC. Ax. 9 .■.ZA0OZ.A0B. Ax. 11 .\ZA0OZ A'O'B', by Ax. 9. q.e.d. This proposition is the converse of Prop. I. 168. Law of Converse Theorems. Of four magnitudes, a, h, x, y, if (1) a~>h when x > y, (2) a = 5 when x = y, and (3) a<h when x<y, then the converses of these three statements are always true. For when a > 6 it is impossible that x — y^ for then a would equal h by (2) ; or that x < y, for then a would be less than 6 by (3). Hence x > y when a > 6. In the same way, x = 2/ when a = h, and x <y when a <b. 169. Chord. A straight line that has its extremi- ^■^^ss-*^ ties on a circle is called a chord. A chord is said to subtend the arcs that it cuts from a circle. Unless the contrary is stated, the chord is taken as subtending: the minor arc. 96 BOOK 11. PLANE GEOMETRY Proposition III. Theorem 170. In the same circle or in equal circles, if two arcs are equal, they are subtended hy equal chords ; and if ttvo arcs are unequal, the greater is subtended by the greater chord. Given two equal circles with centers O and 0\ with arcs AB&nd A'B^ equal, and with arc AF greater than arc A^B\ To prove that 1. chord AB = chord A' B' ; 2. chord AF> chord A' B'. Proof. 1. Draw the radii OA, OB, OF, O'A', O'B'. Since OA = 0'A', and OB = O'B', §162 and ZAOB = ZA'0'B', {In equal d) equal arcs siMend equal central A.) §167 .'. A 0.45 is congruent to A O'A'B', §68 and chord .4 5 = chord .4 '5'. §67 Proof. 2. In the A OAF and O'A'B', OA = 0'A', and OF = O'B', §162 but Z A OF is greater than Z A 'O'B'. §167 (In equal (D, of two unequal arcs the greater subtends the greater central Z. ) . • . chord AF> chord A 'B', by § 115. Q. e. d. 171. Corollary. In the same circle or in equal circles, the greater of two unequal major arcs is subtended hy the less chord. ARCS AND CHORDS 97 Proposition IV. Theorem 172. In the same circle or in equal circles, if two chords are equal, they subtend equal arcs ; and if tivo chords are unequal, the greater subtends the greater arc. Given two equal circles with centers and 0\ with chords AB and A'5' equal, and with chord AF greater than chord A'B\ To prove that 1. arc AB = arc A' B' ; 2. arcAF>arcA'B'. Proof. 1. Draw the radii OA, OB, OF, O'A', O'B'. Since OA = O'A', and OB = O'B', § 162 and chord AB = chord A 'B', Given .'.AOABis congruent to A O'A'B', § 80 and ZAOB = ZA 'O'B'. § 67 .-. arc AB = sltgA'B'. § 166 Proof. 2. In the A OAF and O'A'B', OA = O'A ', and OF = O'B', § 162 but chord ^F> chord ^'5'. Given .\ZAOF>ZA'0'B'. §116 . • . arc ^ jP > arc A 'B', by § 166. Q. e. d. This proposition is the converse of Prop. III. 173. Corollary. In the same circle or in equal circles the greater of two unequal chords subtends the less major arc. 98 BOOK 11. PLANE GEOMETRY Proposition Y. Theorem 174. A line through the center of a circle perpendicular to a chord bisects the chord and the arcs subtended by it. Q Given the line PQ through the center O of the circle AQBP^ perpendicular to the chord AB at M. To prove that AM— BM, arc AQ = arc BQ, and arc AP = arc BP. Proof. Draw the radii OA and OB. Then since OM = OM, Iden. and OA = OB, § 162 .'. rt. A A MO is congruent to rt. A BMO. § 89 .\AM=BM, 3ind ZA0Q = ZQ0B. §67 Likewise Z POA = Z BOP. § 58 . • . arc AQ = arc BQ, and arc AP = arc BP, by § 166. Q. e. d. 175. Corollary 1. A diameter bisects the circle. 176. Corollary 2. A line through the center that bisects a chord is perpendicular to the chord. 177. Corollary 3. The perpendicular bisector of a chord passes through the center of the circle and bisects the arcs subtended by the chord. How many bisectors of the chord are possible ? How many ± bisec- tors ? Therefore with what line must this coincide (§ 174) ? ARCS AND CHORDS 99 Proposition VI. Theorem 178. In the same circle or in equal circles equal chords are equidistant from the center, and chords equidistant from, the center are equal. Given AB and CD, equal chords of the circle ACDB. To prove that AB and CD are equidistant from the. center 0. Proof. Draw OP ±toAB, and OQ _L to CD. Draw the radii OA and OC. OP bisects AB, and OQ bisects CD. § 174 Tlien since AP = CQ, Ax. 4 and OA = OC, § 162 .-. rt. A OP A is congruent to rt. AOQC. § 89 .\OP = OQ. §67 .'. AB and CD are equidistant from 0, by § 88. q.e.d. Given OP and OQ^ equal perpendiculars from the center O to the chords AB and C2). 2^0 prove that AB = CD. Proof. Since OA = OC, § 162 and OP = OQ, Given .-. rt. A OPA is congruent to rt. A OQC. § 89 .'.AP = CQ. §67 .-. .4^ = C/>, by Ax. 3. Q.e.d. 100 BOOK II. PLANE GEOMETRY Proposition VII. Theorem 179. In the same circle or in equal circles, if two chords are unequal, they are unequally distant from the center, and the greater chord is at the less distance. Given a circle with center 0, two unequal chords AB and CD^ AB being the greater, and OP perpendicular to AB^ and OQ per- pendicular to CD. To prove that OP <0Q. Proof. Suppose AE drawn equal to CD, and OP, A-toAE. Draw PR. OP bisects AB, and OR bisects AE. § 174 {A line through the center of a circle ± to a chord bisects the chord.) But AB> CD. Given .-. AB>AE, the equal of CD. Ax. 9 .-. AP>AR. Ax. 6 .'.ZARP>ZRPA. §113 {If two sides of a A are unequal, the A opposite these sides are unequal, and the Z opposite the greater side is the greater.) .'. Z.PRO, the complement of ZARP, is less than Z.OPR, the complement of Z RPA. § 59 .'.OP<OR. §114 But OR=OQ. §178 .-. OP<OQ, by Ax. 9. Q.e.d. AKCS AND CHORDS 101 Proposition VIII. Theorem 180. In the same circle or in equal circles, if tivo chords are unequally distant from the center, they are unequal, and the chord at the less distance is the greater. Given a circle with center 0, two chords AB and CD unequally distant from 0, and OP^ the perpendicular to AB^ less than OQ^ the perpendicular to CD. To prove that AB>CD. Proof. Suppose AE drawn equal to CD, and OR X to AE. Now OP < OQ, Given and OR = OQ. § 178 .-. OP<OR. Ax. 9 Drawing PR, A PRO < A OPR. § 1 1 3 .•. Z ARP, the complement of Z PRO, is greater than Z RPA, the complement of Z OPR. § 59 .\AP>AR. §114 But AP = \AB,2indiAR = \AE. §174 .'.AB>AE. Ax. 6 But CD = AE. Hyp. .*. AB>CD, by Ax. 9. Q.e.d. This proposition is the converse of Prop. VII. 181. Corollary. A diameter of a circle is greater than any other chord. 102 BOOK II. PLANE GEOMETRY 182. Secant. A straight line that intersects a circle is called a secant. In this figure AD is Si secant. Since only two equal obliques can be drawn to a line from an external point (§ 85), and since the two equal angles which radii make (§ 74) with any secant where it cuts the circle cannot be right angles (§ 109), they must be oblique ; and hence it follows that a secant can intersect the circle in only two points. 183. Tangent. A straight line of unlimited length that has one point, and only one, in common with a circle is called a tangent to the circle. In this case the circle is said to be tangent to the line. Thus in the figure, BC is tangent to the circle, and the circle is tangent to BC. The common point is called the point of contact or point of tangency. By the tangent from an external point to a circle is meant the line- segment from the external point to the point of contact. EXERCISE 27 1. A radius that bisects an arc bisects its sub- tending chord and is perpendicular to it. 2. On a circle the point P is equidistant from two radii OA and OB. Prove that P bisects the arc AB. 3. In this circle the chords AM and ]\fB are equal. Prove that M bisects the arc AB and that the radius OM bisects the chord AB. 4. On a circle are five points, A, B, C, D, E, so placed that .45, BC, CD, DE are equal chords. Prove that A C, BD, CE are equal chords, and that AD and BE are also equal chords. 5. If two chords intersect and make equal angles with the diameter through their point of intersec- tion, these chords are equal. SECANTS AND TANGENTS 103 Proposition IX. Theorem 184. A line perpendicular to a radius at its extrem- ity is tangent to the circle. A p Given a circle, with XY perpendicular to the radius OP at P. To prove that XY is tangent to the circle. Proof. From draw any other line to XY, as OA. Then OA>OP. §86 .'. the point A is outside the circle. § 160 Hence every point, except P, of the line AT is outside the circle. Therefore AT is tangent to the circle at P, by § 183. q.e.d. 185. Corollary 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact. For OP is the shortest line from O to XY, and is therefore ± to XY (§ 86) ; that is, XY is ± to OP. 186. Corollary 2. A perpendicular to a tangent at the point of contact passes through the center of the circle. For a radius is ± to a tangent at the point of contact, and therefore a ± erected at the point of contact coincides with this radius and passes through the center of the circle. 187. Corollary 3. A perpendicular from the center of a circle to a tangent passes through the point of contact. What does § 86 say about this perpendicular ? '•■ .- 104 BOOK II. PLANE GEOMETEY 188. Concentric Circles. Two circles that have the same center are said to be concentric. EXERCISE 28 1. The shortest chord that can be drawn through a given point within a circle is that which is perpendicular to the diameter through the poiiit. Show that any other chord, CD, through P, is nearer than is AB. 2. The diameter CD bisects the arc AB. Prove that ZCi^^ =ABAC. What kind of a triangle is A ^ JBC ? 3. Tangents at the extremities of a diameter are parallel. 4. The arc AB is greater than the arc BC. OP and OQ are perpendiculars from the center to AB and BC respectively. Prove that Z QPO is greater than Z OQP. 5. What is the locus of the center of a circle tangent to the line XY dX the point P ? Prove it. What two conditions must be shown to be fulfilled ? 6. What is the locus of the mid-points of a number of par- allel chords of a circle ? Prove it. 7. Three equal chords, AB, BC, CD, are placed /^^T^^^/? end to end, and the radii OA, OB, OC, OD are 2)/ drawn. Prove that AAOC = AB0D. 8. All equal chords of a circle are tangent to a concentric circle. 9. If a number of equal chords are drawn in this circle, the figure gives the impression of a second circle inside the first and concentric with it. Explain the reason. SECANTS AND TANGENTS 105 Proposition^ X. Theorem 189. Two parallel lines intercept equal arcs on a circle. E-~--J!£^- -F A- E Fig. 1 Fig. 2 Fig. 3 Case 1. When the parallels are a tangent and a secant (Fig. 1). Given AB^ a tangent at P, parallel to CD^ a secant. To prove that arc CP = arc DP. Proof. Suppose PP' drawn ± to AB Sit P. Then PP' is a diameter of the circle. § 186 And PP' is also _L to CD. § 97 .-. arc CP = arc Z)P. § 174 Case 2. When the parallels are both secants (Fig. 2). Given AB and CD, parallel secants. To prove that arc AC = arc BD. Proof. Suppose EF II to CD and tangent to the circle at M. Then arc ^3/= arc BM, and arc CM = arc DM. Case 1 .-. arc ylC = arc 57). Ax. 2 Case 3. When the parallels are both tangents (Fig. 3). Given AB^ a tangent at E, parallel to CD^ a tangent at F. To prove that arc FGE = arc FHE. Proof. Suppose a secant GH drawn II to AB. Then arc GE = arc HE, and arc FG = arc FH. Case 1 .'. arc FGE = arc FHE, by Ax. 1. Q. e. d. 106 BOOK II. PLANE GEOMETRY Proposition XI. Theorem 190. TJirough three points not in a straight line one cii'cle, and only one, can he draion. Given -A, J?, C, three points not in a straight line. To prove that one circle, and only one, can he drawn through A, B, and C. Proof. Draw AB and BC. At the mid-points of AB and BC suppose Js erected. These -k will intersect at some point 0, since AB and BC are neither parallel nor in the same straight line. The point O is in the perpendicular bisector of AB, and is therefore equidistant from A and B] the point is also in the perpendicular bisector of BC, and is therefore equidistant from B and C. § 150 Therefore is equidistant from A, B, and C. Therefore a circle described about as a center, with a radius OA, will pass through the three given points. § 160 The center of any circle that passes through the three points must be in both of these perpendicular bisectors, and hence at their intersection. As two straight lines can intersect in only- one point (§ 55), is the only point that can be the center of a circle through the three given points. Q. e. d. 191 . Corollary. Tivo circles can intersect in only two points. If two circles have three points in common, can it be shown that they coincide and form one circle ? SECANTS AND TANGENTS 107 Proposition XII. Theorem 192. Tlie tangents to a circle draivn from an external point are equal, and tnake equal angles ivith the line joining the point to the center. A Given PA and P5, tangents from P to the circle whose center is 0, and PO the line joining P to the center 0. To prove that PA = PB, and ZAPO = Z OPB, Proof. Draw OA and OB. PA is ± to OA, and PB is ± to OB. § 185 {A tangent to a circle is J_ to the radius drawn to the point of contact.) In the rt. A PAO and PBO, PO = PO, Men. and OA = OB. § 162 .-. rt. A PA is congruent to rt. A PBO. § 89 .-. PA = PB, and ZAPO = Z OPB, by § 67. Q. e. d. 193. Line of Centers. The line determined by the centers of two circles is called the line of centers. 194. Tangent Circles. Two circles that are both tangent to the same line at the same point are called tangent circles. Circles are said to be tangent internally or externally, according as they lie on the same side of the tangent Hne or on opposite sides. E.g. the two circles shown in the figure on page 110 are tangent externally. The point of contact with the line is called the point of contact or point of tamjency of the circles. 108 BOOK II. PLANE GEOMETRY EXERCISE 29 1. Show that the reasoning of § 190 will not hold for four points, and hence that a circle cannot always be drawn through four points. 2. Tangents to a circle 3it A, B, C, points on the circle, meet in P and Q, as here shown. Prove that AP -\- QC = PQ. 3. If a quadrilateral has each side tangent to a circle, the sum of one pair of opposite sides equals the sum of the other pair. In this figure, SP -{■ QR = PQ + RS. 4. The hexagon here shown has each side tangent to the circle. Prove that AB-\- CD + EF = BC -^ DE -^ FA. 5. In this figure CF is a diameter perpen- dicular to the parallel chords DB and EA, and arc AB = 40° and arc 5C = 50°. How many de- grees are there in arcs CD, DE, EF, and FA ? 6. In this figure XYis tangent to the circle at B, the chord CA is perpendicular to the j diameter BD, and the arc CD = 150°. How many degrees are there in arc AB? 7. If a quadrilateral has each side tangent to a circle, the sum of the angles at the center subtended by any two opposite sides is equal to a straight angle. S. AP and CQ are parallel tangents meeting a third tangent QP, as shown in the figure. be- ing the center, prove that the angle POQ is a right angle. Are A, 0, and C in the same straight line ? Draw OA and OC, and find the relations of the zi at to those at P and Q. LINE OF CENTERS 109 Proposition XIII. Theorem 195. If tivo circles intersect, the line of centers is the perpendicular bisector of their common chord. Given and 0', the centers of two intersecting circles, AB the common chord, and 00' the line of centers. To prove that 00' is A. to AB at its mid-point Proof. Draw OA, OB, O'A, and O'B. OA = OB, and O'A = O'B. § 162 .'. and O' are two points, each equidistant from A and B. .*. 00' is the perpendicular bisector of AB, by § 151. q.e.d. 196. Common Tangents. A tangent to two circles is called a common external tangent if it does not cut the line-segment joining the centers, and a common internal tangent if it cuts it. EXERCISE 30 Describe the relative position of two circles if the line-segment joining the centers is related to the radii as stated in Exs. 1-5, and illustrate each case hy a figure : 1. The line-segment greater than the sum of the radii. 2. The line-segment equal to the sum of the radii. 3. The line-segment less than the sum but greater than the difference of the radii. 4. The line-segment equal to the difference of the radii. 5. The line-segment less than the difference of the radii. 110 BOOK 11. PLANE GEOMETRY Proposition XIV. Theorem 197. If two circles are tangent to each other, the line of centers passes through the point of contact. A B Given two circles tangent at P. To prove that P is in the line of centers. Proof. Let AB be the common tangent at P. § 194 Then a J_ to ^4 B, drawn through the point P, passes through the centers and 0\ § 186 {A l.to a tangent at the point of contact passes through the center of the circle.) Therefore the line determined by and 0', having two points in common with this _L, must coincide with it. Post. 1 .*. P is in the line of centers. q.e.d. EXERCISE 31 Describe the relative position of two circles having tangents as stated in Exs. 1~5, and illustrate each case by a figure : 1. Two common external and two common internal tangents. 2. Two common external tangents and one common internal tangent. 3. Two common external tangents and no common internal tangent. 4. One common external and no common internal tangent. 5. No common tangent. TANGENTS 111 6. The line which passes through the mid-points of two parallel chords passes through the center of the circle. 7. If two circles are tangent externally, the tangents to them from any point of the common internal tangent are equal. 8. If two circles tangent externally are tangent to a line AB at A and B, their common internal tangent bisects AB. 9. The line drawn from the center of a circle to the point of intersection of two tangents is the perpendicular bisector of the chord joining the points of contact. 10. The diameters of two circles are respectively 2.74 in. and 3.48 in. Find the distance between the centers of the circles if they are tangent externally. Find the distance between the centers of the circles if they are tangent internally. 11. Three circles of diameters 4.8 in., 3.6 in., and 4.2 in. are externally tangent, each to the other two. Find the perimeter of the triangle formed by joining the centers. 12. A circle of center and radius r' rolls around a fixed circle of radius r. What is the locus of O? Prove it. 13. The line drawn from the mid-point of a chord to the mid-point of its subtended arc is perpendicular to the chord. 14. If two circles tangent externally at P are tangent to a line AB at A and B, the angle BPA is a right angle. 15. Three circles are tangent externally at the points A, B, and C, and the chords AB and AC are produced to cut the circle BC Sit D and E. Prove that DE is a diameter. 16. If two radii of a circle, at right angles to each other, when produced are cut by a tangent to the circle at A and B, the other tangents from A and B are parallel to each other. 17. If two common external tangents or two common inter- nal tangents are drawn to two circles, the line-segments inter- cepted between the points of contact are equal. 112 BOOK 11. PLANE GEOMETRY 198. Measure. The number of times a quantity of any kind contains a known unit of the same kind, expressed in terms of that known imit, is called the measure of the quantity. Thus we measure the length of a schoolroom by finding the number of times it contains a known unit called the foot. We measure the area of the floor by finding the number of times it contains a known unit called the square foot. You measure your weight by finding the number of times it contains a known unit called the pound. Thus the measure of the length of a room may be 30 ft., the measure of the area of the floor may be 600 sq. ft., and so on. The abstract number found in measuring a quantity is called its numerical tneasui^e, or usually simply its measure. 199. Ratio. The quotient of the numerical measures of two quantities, expressed in terms of a common unit, is called the ratio of the quantities. Thus, if a room is 20 ft. by 35 ft., the ratio of the width to the length is 20 ft, -=- 35 ft., or ||, which reduces to ^. Here the common unit is 1 ft. The ratio of a to 6 is written -, or a : 6, as in arithmetic and algebra. b Thus the ratio of 20° to 30° is f§, or |, or 2 : 3. 200. Commensurable Magnitudes. Two quantities of the same kind that can both be expressed in integers in terms of a com- mon unit are said to be commensurable magnitudes. Thus 20 ft. and 35 ft. are expressed in integers (20 and 35) in terms of a common unit (1 ft.); similarly 2 ft. and 3| ft., the integers being 4 and 7, and the common unit being \ ft. The common unit used in measuring two or more commensurable magnitudes is called their common measure. Each of the magnitudes is called a multiple of this common measure. 201. Incommensurable Magnitudes. Two quantities of the same kind that cannot both be expressed in integers in terms of a common unit are said to be incommensurable magnitudes. Thus, if a = V2 and 6 = 3, there js no number that is contained an integral number of times in both V2 and 3. Hence a and h are, in this case, incommensurable magnitudes. measureme:n^t 113 202. Incommensurable Ratio. The ratio of two incommensur- able magnitudes is called an incommensurable ratio. Although the exact value of such a ratio cannot be expressed by an integer, a common fraction, or a decimal fraction of a limited number of places, it may be expressed approximately. Thus suppose - = V2. Now V2 =1.41421356 •••, which is greater than 1.414213 but less than 1.414214. Then if a millionth part of b is taken as the unit of measure, the value oi a:b lies between 1.414213 and 1.414214, and therefore differs from either by less than 0.000001. By carrying the decimal further an approximate value may be found that will differ from the ratio by less than a billionth, a trilllonth, or any other assigned value. That is, for i^ractical purposes all ratios are commensurable. For example, if - > — but < , then the error in taking either of 1)71 n these values for - is less than - , the difference of these ratios. But by b 1 "" increasing n indefinitely, - can be decreased indefinitely, and a value of the ratio can be found within any required degree of accuracy. EXERCISE 32 Find a common measure of : 1. 32 in., 24 in. 3. 5^ in., 3^ in. 5. 6^ da., 2| da. 2. 48 ft, 18 ft. 4. 2| lb., 1^ lb. 6. 14.4 in., 1.2 in. Find the greatest common measure of: 7. 64 yd., 24 yd. 9. 7.5 in., 1.25 in. 11. 2| ft, 0.25 ft 8. 51 ft., 17 ft 10. 31 in., 0.33J in. 12. 75°, 7° 30'. 13. lia:b = V3, find an approximate value of this ratio that shall differ from the true value by less than 0.001. 114 BOOK II. PLANE GEOMETKY 203. Constant and Variable. A quantity regarded as having a fixed value throughout a given discussion is called a constant, but a quantity regarded as having different successive values is called a variable. 204. Limit. When a variable approaches a constant in such a way that the difference between the two may become and remain less than any assigned positive quantity, however small, the constant is called the Ihnit of the variable. Variables can sometimes reach their limits and sometimes not. E.g. a chord may increase in length up to a certain limit, the diameter, and it can reach this limit and still be a chord ; it may decrease, approaching the limit 0, but it cannot reach this limit and still be a chord. 205. Inscribed and Circumscribed Polygons. If the sides of a polygon are all chords of a circle, the polygon is said to be inscribed in the circle ; if the sides are all tangents to a circle, the polygon is said to be circum- scribed about the circle. The circle is said to be circum- scribed about the inscribed polygon, and to be inscribed in the circum- scribed polygon. Inscribed Polygon Circumscribed Polygon 206. Circle as a Limit. If we inscribe a square in a circle, and then inscribe an octagon by taking the mid-points of the four equal arcs for the new vertices, the octa- gon is greater than the square but smaller than the area inclosed by the circle, and the perim- eter of the octagon is greater than the perim- eter of the square (§ 112). By continually doubling the number of sides in this way it appears that the area inclosed by the circle is the limit of the area of the polygon, and the circle is the limit of its perimeter, as the number of sides is indefinitely increased. Hence we have limiting forms as well as limiting values, the form of the circle being the limit approached by the form of the inscribed polygon. LIMITS 115 -\L ■\M 207. Principle of Limits. If, ivhile approaching their respec- tive limits, two variables are always equal, their limits are equal. Let AX and BY increase in length in such a way that they always remain equal, and let their respective limits be ylL and BM. To prove that AL = BM. Suppose these limits are not equal, but that AZ = BM. Then since X may reach a point between Z and L we may have AX> AZ, and therefore greater than its supposed equal, BM; but 5 F cannot be greater than BM. Therefore we should have AX> BY, which is contrary to what is given. Hence BM cannot be greater than AL, and similarly AL cannot be greater than BM. .'. AL = BM. q.e.d. 208. Area of Circle. The area inclosed by a circle is called the area of the circle. It is evident that a diameter bisects the area of a circle. 209. Segment. A portion of a plane bounded by an arc of a circle and its chord is called a segment of the circle. If the chord is a diameter, the segment is called a semicircle, this word being commonly used to mean not only half of the circle but also the area inclosed by a semicircle and a diameter. 210. Sector. A portion of a plane -^m,c.b- bounded by two radii and the arc of the circle intercepted by the radii is called a sector. If the arc is a quarter of the circle, the sector is called a quadrant. 211. Inscribed Angle. An angle whose vertex is on a circle, and whose sides are chords, is called an inscribed angle. An angle is said to be inscribed in a segment if its vertex is on the arc of the segment and its sides pass through the ends of the arc. 116 BOOK II. PLANE GEOMETRY Proposition XV. Theorem 212. In the same circle or in equal circles tivo central angles have the same ratio as their intercepted arcs. Fig. 1 Fig. 2 Fig. 3 Given two equal circles with centers and 0\ AOB and A'O'B' being central angles, and AB and A^B' the intercepted arcs. To prove that , , ^ — = Case 1. When the ares are commensurahle (Figs. 1 and 2). Proof. Let the arc m be a common measure of A^B' and AB. Apply the arc m as a measure to the arcs A'B^ and ^5 as many times as they will contain it. Suppose m is contained a times in A'B', and b times in AB. Then arc A *B' a .SiVcAB b At the several points of division on AB and A 'B' draw radii. These radii will divide Z.AOB into b j)arts, and Z.A'0'B' into a parts, equal each to each. § 167 ZA'O'B' a ZAOB ~1j' ZA'O'B' slygA'B' , , „ .-. ■ ^ ^^ = — J by Ax. 8. Q.E.D. ZAOB SiYcAB ^ Case 2 may be omitted at the discretion of the teacher if the incom- mensurable cases are not to be taken in the course. MEASURE OF ANGLES 117 Case 2. When the arcs are incommensurable (Figs. 2 and 3). Proof. Divide AB into a number of equal parts, and apply one of these parts to A 'B' as many times as A 'B' will contain it. Since AB and A'B' are incommensurable, a certain number of these parts will extend from A' to some point, as P, leaving a remainder PB' less than one of these parts. Draw O'P. By construction AB and A'P are commensurable. ZA'O'P arc^'P * ' ZAOB ^vgAB Case 1 By increasing the number of equal parts into which AB is divided we can diminish the length of each, and therefore can make PB' less than any assigned positive value, however small. Hence PB' approaches zero as a limit as the number of parts of AB is indefinitely increased, and at the same time the corresponding angle PO'B' approaches zero as a limit. § 204 Therefore the arc ^'P approaches the arc A'B' as a limit, and the A A' O'P approaches the Z A'O'B' as a limit. . , , arc^'P , arcyl'5' .. .^ .*. the variable -— approaches — — as a Innit, arc^^ arc .45 • VI AA'O'P , Z. A'O'B' .. .. and the variable , , ,,„ approaches . . __ as a limit. Z.AOB ^^ AAOB ZA'O'P . , , ^ arc^'P But . , ^,, IS always equal to -— > Z.AOB ./ ^ siTcAB as A'P varies in value and approaches A'B' as a limit. Case 1 Z A'O'B' SiYcA'B' , . OAT .^T^ .'.-^ = --— jby§207. Q.E.D. ZAOB SiTcAB -^ 213. Numerical Measure. We therefore see that the numerical measure of a central angle (in degrees, for example) equals the numerical measure of the intercepted arc. This is commonly expressed by saying that a central angle is measured by the intercepted arc. 118 BOOK 11. PLANE GEOMETRY Proposition XVI. Theorem 214. A7i inscribed angle is measured hy half the in- tercepted arc. B B B Given a circle with center O and the inscribed angle B^ inter- cepting the arc AC. To prove that Z. B is measured hy half the arc A C. Case 1. When is on one side, as AB (Fig. 1). Proof. Draw OC. Then '.'OC = OB, §162 .\AB = AC. §74 But Z.B-\-AC = Z. A OC. § 111 .•.2AB = AA0C. Ax. 9 .'.Z.B = :^ Z.AOC. Ax. 4 But AAOC is measured by arc A C. § 213 .*. 1 Z. AOC is measured by ^ arc AC. Ax. 4 .'. Z.B is measured by ^ arc AC. Ax. 9 Case 2. When lies within the angle B (Fig. 2). Proof. Draw the diameter BD, Then Z ABD is measured by \ arc AD, and Z DBC is measured by ^ arc DC. Case 1 .-. Z ^57) + Z Z)5C is measured by \ (arc ^i) + arc i)C), or Z ^^C is measured by \ arc AC. MEASURE OF ANGLES 119 Case 3. W?ien lies outside the angle B (Fig. 3). Proof. Draw the diameter BD. Then Z DBC is measured by \ arc 7)C, and Z DBA is measured by \ arc DA. Case 1 .'. /.DBC — Z. DBA is measured by \ (arc DC — arc /).!), or Z.ABC is measured by \ arc .4C. Q.E.D. Fig. 4 Fig. 6 215. Corollary 1. An angle inscribed in a semicircle is a right angle. For it is half of a central straight angle, as,in Fig, 4. 216. Corollary 2. An angle inscribed in a segment greater than a semicircle is an acute angle, and an angle inscribed in a segment less than a semicircle is an obtuse angle. See A A and B in Fig, 5. 217. Corollary 3. Angles inscribed in the same segment or in equal segments are equal. Why is this ? (Fig. 6.) 218. Corollary A:. If a quadrilateral is inscribed in a circle, the opposite angles are supplementary ; and, conversely, if two opposite angles of a quadrilateral are supplementary, the quadrilateral can be inscribed in a circle. For the second part, can a circle be passed through A^ /f^f/^X jB, C (§ 190) ? If it does not pass through D also, can you f I d>--~^A show that Z D would be greater than or less than some other jj^^^^^"^ angle (§111) that is supplementary to Z S ? 120 BOOK II. PLAXE GEOMETRY EXERCISE 33 1. A parallelogram inscribed in a circle is a rectangle. 2. A trapezoid inscribed in a circle is isosceles. 3. The shorter segment of the diameter through a given point within a circle is the shortest line that can be drawn from that point to the circle. f o p^ Let P be the given point. Prove PA shorter than any other line PX from P to the circle. 4. The longer segment of the diameter through a given point within a circle is the longest line that can be drawn from that point to the circle. 5. The diameter of the circle inscribed in a right triangle is equal to the difference between the hypotenuse and the sum of the other two sides. 6. A line from a given point outside a circle passing through the center contains the shortest line-segment that can be drawn from that point to the circle. Let P be the point, the center, A the point where PO cuts the circle, and C any other point on ^' the circle. How does PC\-CO compare with P02 . 7. A line from a given point outside a circle passing through the center contains the longest line-segment (to the concave arc) that can be drawn from that point to the circle. 8. Through one of the points of intersection of two circles a diameter of each circle is drawn. Prove that the line joining the ends of the diameters passes through the other point of intersection. 9. If two circles intersect and a line is drawn through each point of intersection terminated by the circles, the chords joining the corresponding ends of these lines are parallel. MEASURE OF ANGLES 121 Proposition XVIT. Theorem 219. A71 angle formed hy two chords intersecting ivithin the circle is measured hy half the sum of the intercepted arcs. Given the angle AOB formed by the chords AC and BD. To prove that Z A OB is measured hy ^ (ar<? AB -f- arc CD^. Proof. ' Draw A D. Then AAOB = AA+Z.D. §111 {An exterJ,or Z 0/ a A is equal to the sum of the two opposite interior A.) But Z A is measured by ^ arc CD, § 214 {An inscribed Z is measured by half the intercepted arc.) and Z Z) is measured by ^ arc AB. § 214 .-. Z. AOB is measured by J (arc A B + arc CD), by Ax. 1. Q. e. d. Discussion. If is at the center of the circle, to what previous prop- osition does this proposition reduce ? If is on the circle, as at U, to what previous proposition does this proposition reduce ? Suppose the point remains as in the figure, and the chord AC swings ahout as a pivot until it coincides with the chord BD. What can then be said of the measure of A AOB and COD ? What can be said as to the measure oi ABOC and DO A ? It is also possible to prove the proposition by drawing a chord AE parallel to BD, and showing that ZAOB = ZA, since they are alternate- interior angles formed by a transversal cutting two parallels. Now Z A is measured by I arc CE. But arc CE = arc CD + arc DE^ or arc CD + arc ^B, since arc AB = arc DE (§ 189). Therefore ZAOB, which equals ZA, is measured by i (arc AB -\- arc CD). 122 BOOK II. PLAKE GEOMETRY Propositiox XVIII. Theorem 220. An angle formed hy a tangent and a chord drawn from the point of contact is measured hy half the in- tercepted arc. Given the chord PQ and the tangent XY through P. To prove that A QPX is measured hy half the are QSP. Proof. Suppose the chord QPi, drawn from the point Q par- allel to the tangent XY. Then arc PA' = arc QSP. * § 189 {Two parallel lines intercept equal arcs on a circle.) Also Z QPX = Z PQR. § 100 {If two parallel lines are cut by a transversal, the alternate-interior angles are equal.) But Z PQR is measured by J arc PR. § 214 {An inscribed Z is measured by half the intercepted arc.) Substitute Z QPX for its equal, the Z PQR, and substitute arc QSP for its equal, the arc PR. Then Z QPX is measured by J arc QSP, by Ax. 9. q.e.d. Discussion. By half of what arc is Z FPQ, the supplement of Z QPX, measured ? If PQ should be drawn so as to be perpendicular to XY, by what would A YPQ and QPX be measured ? Suppose PQ swings about the point P as a pivot until it coincides with XY, by what will Z YPQ be measured ? By what will Z QPX be measured, and what will it equal ? MEASURE OF ANGLES 123 Proposition XIX. Theorem 221. An angle formed hy two secants, a secant and a tangent, or tivo tangents, drawn to a circle from an external point, is measured hy half the difference of the intercepted arcs. Fig. 1 Fig. 3 Given two secants PBA and PCD^ from the external point P. To prove that Z P is measured hy ^ (arc DA — arc BC), Proof. Suppose the chord BX drawn II to PCD (Fig. 1). Then Furthermore §189 Ax. 9 §102 §214 arc BC = arc DX. arc XA = arc DA — arc DX. .'. arc XA — arc DA — arc BC. Also ZP = ZXBA. But Z XBA is measured by ^ arc XA. Substitute AP for its equal, the Z.XBA, and substitute arc DA — arc BC for its equal, the arc XA. Then Z P is measured by J (arc DA — arc BC), hy Ax. 9. Q. e. d. If the secant PBA Y swings around to tangency, it becomes the tangent PB and Fig. 1 becomes Fig. 2. If the secant PCD also swings around to tangency, it becomes the tangent PC and Fig. 2 becomes Fig. 3. The proof of the theorem for each of these cases is left for the student. 124 BOOK 11. PLANE GEOMETRY EXERCISE 34 1. If two circles touch each other and two lines are drawn through the point of contact terminated by the circles, the chords joining the ends of these lines are parallel. This could be proved if it could be shown that Z A equals what angle ? To what two angles can these angles be proved equal by § 220 ? Are those angles equal ? 2. If one side of a right triangle is the diameter of a circle, the tangent at the point where the circle cuts the hypotenuse bisects the other side. If OE is II to J.C, then because BO = OA, what is the relation of BE to EC ? The proposition therefore reduces to proving that OE is parallel to what line ? This can be proved if Z BOE can be shown equal to what angle ? 3. If from the extremities of a diameter AD two chords, A C and DB, are drawn intersecting at P so as to make Z APB = 45°, then Z BOC ^ is a right angle. 4. The radius of the circle inscribed in an equilateral tri- angle is equal to one third the altitude of the ^ triangle. To prove this we must show that AF equals what line? It looks as if AF might equal EF, and EF equal OF. Is there any way of proving A OFE equi- lateral ? of proving A AEF isosceles ? 5. If two lines are drawn through any point in a diagonal of a square parallel to the sides, the points where these lines meet the sides lie on the circle whose center is the point of intersection of the diagonals. OY = OZ if what two A are congruent? Why are these A congruent ? OY = OX if what two A are con- gruent ? OX = 0W if what two A are congruent ? a PRINCIPLE OF CONTINUITY 125 222. Positive and Negative Quantities. In geometry, as in algebra, quantities may be distinguished as positive and as negative. Thus as we consider temperature above zero posi- tive and temperature below zero negative, so in this figure, if OB is considered positive, then OD may be considered negative. Similarly, if OA is considered positive, then OC may be considered negative. Likewise with respect to angles and arcs, if the rotating line OA moves in the direction of AB, counterclockwise, the angle and arc generated are considered positive. If it rotates in the direction AB', like the hands of a clock, the angle and arc generated are considered negative. 223. Principle of Continuity. By considering the distinction between positive and negative magnitudes, a theorem may often be so stated as to include several particular theorems. For example. The angle included between two lines that cut or are tangent to a circle is measured by half the sum of the intercepted arcs. In particular : 1. If the lines intersect at the center, half the sum of the arcs will then become simply one of the arcs, and the proposition reduces to that of § 213. 2. If the lines are two general chords, we have the case of § 219. 3. If the point of in- tersection P moves to the circle, we have the case ( \/ \ f 2\ \ i /P\ of § 214, one arc becom- ing zero. 4. If P moves outside the circle, then the smaller arc passes through zero and becomes negative, so that the sum of the arcs becomes their arithmetical difference (§ 221). We may continue the discussion so as to include all the cases of the propositions proved from § 213 to § 221. When the reasoning employed to prove a theorem is con- tinued as just illustrated, so as to include several theorems, we are said to reason by the Principle of Continuity. 126 BOOK II. PLANE GEOMETRY 224. Problems of Construction. At the beginning of the study of geometry some directions were given for simple construc- tions, so that figures might be drawn with accuracy. It was not proved at that time that these constructions were correct, because no theorems had been studied on which proofs could be based. It is now purposed to review these constructions, to prove that they are correct, and to apply the methods employed to the solution of more difficult problems. 225. Nature of a Solution. A solution of a problem has one requirement that a proof of a theorem does not have. In a theorem we have three general steps : (1) Given, (2) To prove, (3) Proof. In a problem we have four steps : (1) Given, (2) Required (to do some definite thing), (3) Construction (show- ing how to do it), (4) Proof (that the construction is correct). We prove a theorem, but we solve a problem, and then prove that our solution is a correct one. In the figures of this text given lines are shown as full, black lines ; construction lines and lines required are shown as dotted lines. 226. Discussion of a Problem. Besides the four necessary general steps in treating a problem, there is a desirable step to be taken in many cases. This is the discussion of the prob- lem, in which is considered whether there is more than one solution, and other similar questions. For example, suppose the problem is this : Required from a given point to draw a tangent to a circle. After the problem has been solved we may discuss it thus : In general, if the given point is outside the circle, two tangents may be drawn, and these tangents are equal (§ 192) ; if the given point is on the circle only one tangent can be drawn, since only one perpendicular can be drawn to a radius at its extremity (§ 184) ; if the given point is within the circle, evidently no tangent can be drawn. In the discussion the Principle of Continuity often enters, the figure being studied for various positions of some given point or line, as was done in the discussions on pages 121 and 122. PKOBLEMS OF CONSTRUCTION 127 Proposition XX. Problem 227. To let fall a perpendicular iq^on a given line from a given external point. P\ Yy Given the line AB and the external point P. Required from P to let fall a J_ upon AB. Construction. With P as a center, and a radius sufficiently great, describe an arc cutting ^1^ at J^ and Y. Post. 4 With X and Y as centers, and a convenient radius, describe two arcs intersecting at C. Post. 4 Draw PC. Post. 1 Produce PC to intersect AB at M. Post. 2 Then PM is the line required. Q. e. f. Proof. Since P and C are by construction two points each equidistant from X and F, they determine the perpendicular to Jl F at its mid-point. § 151 {Two points each equidistant from the extremities of a line determine the ± bisector of the line.) Q. E. D. Discussion. The following are interesting considerations : That PC produced will really intersect AB, as stated in the construc- tion, is shown in the proof. A convenient radius to take for the two intersecting arcs is XY. If C falls on P, take C at the other intersection of the arcs below AB^ as is seen in the figure of Ex. 2, p, 9. To obtain a radius for the first circle, draw any line from P that will cut AB, and use that. 128 BOOK II. PLANE GEOMETRY Proposition XXI. Problem 228. At a given point in a given line, to erect a per- pendicular to the line, •>.a- — -. \y I ! \ Oy / -^' ^ \X P Yl " X--^^. Fig. 1 Fig. 2 Given the point P in the line AB. Required to erect a A. to AB at P. Case 1. When the point P is not at the end of AB (Fig. 1). Construction. Take PX = PY. Post. 4 With X and Y as centers, and a convenient radius, describe arcs intersecting at C. Post. 4 Draw CP. Post. 1 Then CP is ±to AB. q.e.f. Proof. P and C, two points each equidistant from X and Y, determine the ± bisector of ZF, by § 151. q.e.d. Case 2. When the point P is at the end of AB (Fig. 2). Construction. Suppose P to coincide with B. Take any point outside of AB, and with as a center and OB as a radius describe a circle intersecting AB a,t X. From X draw the diameter XY, and draw BY. Post. 1 Then i^F is ± to AB. Q.e.f. Proof. Z 5 is a right angle. § 215 .-. BY is ± to AB, by § 27. Q.e.d. Discussion. If the circle described with as a center is tangent to AB ?it B, then OB is the required perpendicular (§185). PROBLEMS OF CONSTRUCTION Proposition XXII. Problem 229. To bisect a given line. I 129 ■B M Given the line AB. Required to bisect AB. Construction. With A and B as centers and AB as a radius describe arcs intersecting at X and Y, and draw XY. Post. 4 Then XY bisects AB. q.e.f. Proof. XY bisects AB, by § 151. q.e.d. Proposition XXIII. Problem 230. To bisect a given arc. \g Given the arc AB. Required to bisect AB. Construction. Draw the chord AB. Post. 1 Draw CM, the perpendicular bisector of the chord AB. § 229 Proof. Then CM bisects the arc AB. CM bisects the arc AB, by § 177. Q.E.F. Q.E.D. 130 BOOK 11. PLANE GEOMETRY Proposition XXIV. Problem 231. To bisect a given angle. o Given the angle AOB. Required to bisect A AOB. Construction. With O as a center and any radius describe an arc cutting OA at X and OB at Y. Post. 4 With X and Y as centers and .Y F as a radius describe arcs intersecting at P. Post. 4 Draw OP. Post. 1 Then OP bisects Z.AOB. q.e.f. Proof. Draw PX and PY. Then prove that the A OXP and YP are congruent. § 80 Then AA0P = Z.P0B,hj4 67. Q. e. d. EXERCISE 35 1. To construct an angle of 45°; of 135°. 2. To construct an angle of 22° 30'; of 157° 30'. 3. To construct an equilateral triangle, having given one side, and thus to construct an angle of 60°. 4. To construct an angle of 30° ; and thus to trisect a right angle. 5. To construct an angle of 15°; of 7° 30'; of 195°; of 345°. 6. To construct a triangle having two of its angles equal to 75°. Is the triangle definitely determined ? PROBLEMS OF CONSTRUCTIOISr 131 Proposition XXV. Problem 232. From, a given point in a given line, to draw a line making an angle equal to a given angle. p- Ic \m * Given the angle AOB and the point P in the line PQ. Required from P to draw a line making with the line PQ an angle equal to A A OB. Construction. With O as a center and any radius describe an arc cutting OA at C and OB at D. Post. 4 With P as a center and the same radius describe an arc MX^ cutting PQ at M, Post. 4 With M as a center and a line joining C and D as a radius describe an arc cutting the arc MX at N. Post. 4 Draw PN. Post. 1 Then Z QPN = Z A OB. Q. e. f. Proof. Draw CD and MN. Then prove that the A PMN and OCD are congruent. § 80 Then Z QPN =Z.A OB, by § 67. Q. e. d. 233. Corollary. Through a given external point, to draiv a line parallel to a given line. X Let AB be the given hne and P the given external „/' point. C -f-^-D Draw any Une XPY through P, cutting AB as in /q the figure. / Draw CD through P, making Z.p = Zq. / The Hne CD will be the line required. 132 BOOK II. PLANE GEOMETRY Proposition XXVI. Problem 234. To divide a given line iiito a given number of equal parts. Ai^ 1 ; iB /^ 7 7 7 ' ^^-. / Given the line AB. Required to divide AB into a given number of equal parts. Construction. From A draw the line AO, making any con- venient angle with AB. Post. 1 Take any convenient length, and by describing arcs apply it to ^ as many times as is indicated by the number of parts into which ^^ is to be divided. Post. 4 From C, the last point thus found on A 0, draw CB. Post. 1 From the division points on ^ draw parallels to CB. § 233 These lines divide AB into equal parts. q.e.f. Proof. These lines divide AB into equal parts, by § 134. q. e. d. EXERCISE 36 1. To divide a given line into four equal parts. 2. To construct an equilateral triangle, given the perimeter. 3. Through a given point, to draw a line which shall make equal angles with the two sides of a given angle. 4. Through a given point, to draw two lines so that they shall form with two intersecting lines two isosceles triangles. 5. To construct a triangle having its three angles respec- tively equal to the three angles of a given triangle. PKOBLEMS OF CONSTRUCTION 133 Proposition XXVII. Problem 235. To construct a triangle lohen two sides and the included angle are given. 4. :i— X ;c B Given h and c two sides of a triangle, and O the included angle. Required to construct the triangle. Construction. On any line as AX, by describing an arc, mark o^ AB equal to c. Post. 4 At A construct A BAD equal to Z 0. § 232 On AD, by describing an arc, mark off ^ C equal to h. Post. 4 Draw BC. Post. 1 Then A ABC is the A required. q.e.f. Proof. (Left for the student.) 236. Corollary 1. To construct a triangle when a side and two angles are given. There are two cases to be considered: (1) when the given side is included between the given angles ; and (2) when it is not (in which case find the other angle by § 107). 237. Corollary 2. To construct a triangle when the three sides are given. 238. Corollary 3. To construct a parallelogram when two sides and the included angle are given. Combine § 235 and § 233. 134 BOOK II. PLANE GEOMETRY Proposition XXVIII. Problem 239. To construct a triangle ivheji two sides and the angle opposite one of them are given. yY O/' Given a and h two sides of a triangle, and A the angle opposite a. Required to construct the triangle. Construction. Case 1. If a is less than h. Construct Z XA Y equal to the given A A. § 232 On A Y take A C equal to h. Prom C as a center, with a radius equal to a, describe an arc intersecting the line AX dX B and B\ Draw EC and B'C, thus completing the triangle. Then both the ^ABC and AB^C satisfy the conditions, and hence we have two constructions. q.e.f. This is called the ambiguous case. Discussion. If the given side a is equal to the ± CB, the arc described from C will yi touch AX, and there will be but one con- struction, the rt. A ABC. If the given side a is less than the per- pendicular from C, the arc described from ^ C will not intersect or touch AX, and hence yY a construction is impossible. (j / If Z A is right or obtuse, a construe- y^ \ ^ tion is impossible, since a<h\ for the side X \ ^ of a triangle opposite a right or obtuse angle X is the longest side (§114). A .Y PROBLEMS OF CONSTRUCTION 135 Case 2. If a is equal to b. If the given Z. A is acute, and a = b, the arc described from C as a center, and with a radius equal to a, will cut the line WX at the points A and B. y There is therefore but one triangle that ?<^ satisfies the conditions, namely the isos- ^^ :^ll'____ \/ _ v- celes A ABC. ^^^- — -''^ Discussion. If the ZA is right or obtuse, a construction is impossible when a = b ; for equal sides of a triangle have equal angles opposite them, and a triangle cannot have two right angles or two obtuse angles (§ 109). Case 3. If a is greater than h. If the given Z A is acute, the arc described from C will cut the line WX on opposite sides of .1, at B and B'. The A ABC satisfies the conditions, but the A AB'C does not, for it does not contain the acute Z.A. C "^ There is then only one triangle that • a^y/^\a ] satisfies the conditions, namely the ^.-.l\grS/. ^:>J...x A ABC. ^'^^< '^^'^ If the given Z .1 is right, the arc described from C cuts the line WX on opposite sides of A A at the points B and B', and we have the x^ / j^>\ / two congruent right triangles ABC and AB'C ^^ "b^"^~~"^'b''^ that satisfy the conditions. If the given Z .1 is obtuse, the arc de- \^ scribed from C cuts the line WX on , n/ \\a , oj)posite sides of A, at the points B and -^y i^'l i..i^i.-x B'. The A ABC satisfies the conditions, but the A AB'C does not, for it does not contain the obtuse Z.A. There is then only one triangle that satisfies the con- ditions, namely the A ABC, Discussion. We therefore see that when a > 6, we have only one triangle that satisfies the conditions, for the two congruent right tri- angles give us only one distinct triangle. 136 BOOK 11. PLANE GEOMETRY Proposition XXIX. Problem 240. To circumscribe a circle about a given triangle. ^^ ^c Given the triangle ABC. Required to circumscribe a O about A ABC. Construction. Draw the perpendicular bisectors of the sides ^^and^C. §229 Since AB is not the prolongation of CA, these Js will inter- sect at some point 0. Otherwise they would be II, and one of them would have to be ±. to two intersecting lines. § 82 With as a center, and a radius OA, describe a circle. Post. 4 The O^^C is the O required. q.e.f. Proof. The point is equidistant from A and B, and also is equidistant from A and C. § 150 .*. the point is equidistant from A, B, and C. .*. a O described with as a center, with a radius equal to OA, will pass through the vertices A, B, and C, by § 160. q.e.d. 241. Corollary 1. To describe a circle through three points not in the same straight line. 242. Corollary 2. To find the center of a given circle or of the circle of which an arc is given. 243. Circumcenter. The center of the circle circumscribed about a polygon is called the circumcenter of the polygon. PROBLEMS OF CONSTEUCTION Proposition XXX. Problem 244. To inscribe a circle in a given triangle. 137 P B Given the triangle ABC. Required to inscribe a O in A ABC. Construction. Bisect the A A and B. § 231 From 0, the intersection of the bisectors, draw OP 1. to the side AB. § 227 With as a center and a radius OP, describe the O PQR. The OPQR is the O required. q.e.f. Proof. Since is ii> the bisector of the Z.A,it is equidistant from the sides AB and AC; and since is in the bisector of the Z.B, it is equidistant from the sides AB and BC. § 152 .'.a circle described with as a center, and a radius OP, will touch the sides of the triangle, by § 184. q.e.d. 245. Incenters and Excenters. The center of a circle inscribed in a polygon is called the incenter of the polygon. The intersections of the bisectors of the exterior angles of a triangle are the centers of three circles, each tangent to one side of the triangle and the two other sides produced. These three circles are called escribed circles ; and their centers are called the excenters of the triangle. 138 BOOK II. PLANE GEOMETRY Proposition XXXI. Problem 246. Through a given. point, to draw a tangent to a given circle. . — .P --^ M,--' -^''.P Fig. 1 Given the point P and the circle with center 0. Required through P to draw a tangent to the circle. Case 1. When the given point is on the circle (Fig. 1). Construction. From the center draw the radius OP. Post. 1 Through P draw AT _L to OP. § 228 Then AF is the tangent required. q.e.f. Proof. Since AF is ± to the radius OP, Const. .*. AF is tangent to the O at P, by § 184. q.e.d. Case 2. When the given point is outside the circle (Fig. 2). Construction. . Draw OP. Post. 1 Bisect OP. § 229 With the mid-point of OP as a center and a radius equal to I" OP, describe a circle intersecting the given circle at the points M and N, and draw PM. Then PM is the tangent required. Q. e. f. Proof. Draw OM. Z OMP is a right angle. § 215 .'.PilfisJ- to OJ/. !27 .*. PM is tangent to the circle at M, by § 184. q.e.d. Discussion. In like manner, we may prove PN tangent to the given O. PEOBLEMS OF CONSTRUCTION 139 Proposition XXXII. Problem 247. Upon a given line as a chord, to describe a segment of a circle in which a given angle may he inscribed. ,' /o \ \ / \ / ! ""v '. ly Given the line AB and the angle m. Required on AB as a chords to describe a segment of a circle in which Z. m may be inscribed. Construction. Construct the ZABX equal to the Z.m. § 232 Bisect the line AB by the ± PO. § 229 From the point B draw BO _L to XB. § 228 • With 0, the point of intersection of PO and BO, as a center, and a radius equal to OB, describe a circle. The segment ABQ is the segment required. q.e.f. Proof. The point is equidistant from A and B. § 150 .'. the circle will pass through A and B. § 160 But BX is ± to OB. Const. .'. BX is tangent to the O. § 184 .'. AABX is measured by ^ arc ^5. § 220 But any angle, as the Z Q, inscribed in the segment ABQ is measured by ^ a,vcAB. § 214 .\ZQ = ZABX. Ax. 8 But Z.ABX = Zm. Const. .'. Z.m may be inscribed in the segment ABQ, by § 217. Q. e.d. 140 BOOK 11. PLANE GEOMETRY 248. How to attack a Problem. There are three common methods by which to attack a new problem : (1) By synthesis ; (2) By analysis ; (3) By the intersection of loci. 249. Synthetic Method. If a problem is so simple that the solution is obvious from a known proposition, we have only to make the construction according to the proposition, and then to give the synthetic proof, if a proof is necessary, that the construction is correct. It is rarely the case, however, that a problem is so simple as to allow this method to be used. We therefore commonly resort at once to the second method. 250. Analytic Method. This is the usual method of attack, and is as follows : (1) Suppose the problem solved and see what results follow. (2) Then see if it is possible to attain these results and thus effect the required construction; in other words, try to work backwards. The third method, by the intersection of loci, is considered on page 143. 251. Determinate, Indeterminate, and Impossible Cases. A problem that has a definite number of solutions is said to be determinate. A problem that has an indefinite number of solu- tions is said to be indeterminate. A problem that has no solu- tion is said to be impossible. For example, to construct a triangle, having given its sides, is deter- minate ; to construct a quadrilateral, having given its sides, is indetermi- nate ; to construct a triangle with sides 2 in., 3 in., and 6 in. is impossible. 252. Discussion. The examination of a problem with refer- ence to all possible conditions, particularly with respect to the number of solutions, is called the discussion of the problem. Discussions have been given in several of the preceding problems. SOLUTION OF PKOBLEMS 141 253. Applications of the Anal3rtic Method. The following are examples of the use of analysis in the solution of problems. EXERCISE 37 1. In a triangle ABC, to draw PQ parallel to the base AB, cutting the sides in P and Q, so that PQ shall equal AP -\- BQ. Analysis. Assume the problem solved. Then AP must equal some part of PQ, as PX, and BQ must equal QX. But if AP = PX, what must Z PXA equal ? •.• PQ is II to AB, what does Z PXA equal ? Then why must ZBAX = ZXAP? Similarly, what about ZQBX and Z XBA ? Construction. Now reverse the process. "What should we do to zi yl and B in order to fix X ? Then how shall PQ be drawn ? Now give the proof. 2. To construct a triangle, having given the perimeter, one angle, and the altitude from the vertex of the given angle. Analysis. Let ABC hQ the triangle, Z G the given angle, and CP the given altitude, and assume that the problem is solved. Since the perimeter is given as a definite line, we X- now try producing AB and BA, making BN=BC, and AM=AC. Then Zm = what angle, and Zn = what angle ? Then Zm-\- Zn-\- Z MCN = 180°. But ZMCN=Zm'-^ZACB-\-Zn\ .'. 2 Zm + 2 Zn-^ZACB = 180°. (Why?) .-. Zm + Zn + IZACB = 90°, or Zm-{- Zn = 90°-iZACB. .: Z MCN =90° -\-^ZA CB. (Why ?) .-. Z MCN is known. Construction. Now reverse the process. Draw MN equal to the perim- eter. Then on MN construct a segment in which Z MCN may be inscribed (§ 247). Draw XC II to MN at the distance CP from MN, cutting the arc at C. Then A and B are on the ± bisectors of CM and CN. Why ? PB 142 BOOK II. PLANE GEOMETRY 3. To draw through two sides of a triangle a line parallel to the third side, so that the part intercepted c ^ between the sides shall have a given length. If PQ = d, what does AR equal ? How will you reverse the reasoning ? A R ~b 4. To draw a tangent to a given circle so that it shall be parallel to a given line. 5. To construct a triangle, having given a side, an adjacent angle, and the difference of the other sides. If AB^ ZA, and AC — BC are known, what points are determined ? Then can XB be drawn ? What kind of a tri- -^ angle is A XBC ? How can C be located ? ^ « 6. To construct a triangle, having given two angles and the sum of two sides. ^ Can the third Z be found ? Assume the prob- lem solved. If AX = AB + BC, what kind of a triangle is A BXC ? What does Z CBA equal ? Is Z X known ? How can C be fixed ? 7. To construct a square, having given the diagonal. 8. To draw through a given point P between the sides of an angle AOB ?i line terminated by the sides ^ of the angle and bisected at P. / If PM=: PN, and PR is II to AO, what can you 2?/>_>p say as to OR and RN? Can you now reverse this? / Similarly, if PQ is II to BO, is OQ = to QM? ^ ^ MA 9. To draw a line that would bisect the angle formed by two lines if those lines were produced to meet. li AB and CD are the given lines, consider what would be the con- ditions if they could be produced to meet at 0. Then the q bisector of Z O would be the ± bisector of PQ, a line drawn /\s^ so as to make equal angles with the two given lines. / I \ Now reverse this. How can we draw PQ so as to make / j >^ ZP = ZQ? Draw BR II to DC, and lay off BR = BQ. ^ Then draw QRP and prove that this is such a line. Then draw its _L bisector. EXERCISES IN LOCI 14B 254. Intersection of Loci. The third general method of attack mentioned in § 248 is by intersection of loci. This is very con- venient when we wish to find a point satisfying two conditions, each of which involves some locus. EXERCISE 38 1. To find a point that is i in. from a given point and ^\ in. from a given line. ---^'-— _w_ If P is the given point, what is the locus of a — I- ; ^ B a point I in. from P ? If ^^ is the given line, > ^ l__ what is the locus of a point y\ in. from AB ? --—■-' These two loci intersect in how many points at most? Discuss the solution. 2. To find a point that is J in. from one given point and | in. from another given point. Discuss the number of possible points answering the conditions. 3. To find a point that is \ in. from the vertex of an angle and equidistant from the sides of the angle. 4. To find a point that is equidistant from two intersecting lines and \ in. from their point of intersection. How many such points can always be found ? 5. To find a point that is \ in. from a given point and equi- distant from two intersecting lines. Discuss the problem for various positions of the given point. 6. To find a point that is \ in. from a given point and equi- distant from two parallel lines. Discuss the problem for various positions of the given point. 7. Find the locus of the mid-point of a chord of a given length that can be drawn in a given circle. 8. rind the locus of the mid-point of a chord drawn through a given point within a given circle. U-l G '0\ 144 BOOK II. PLANE GEOMETRY 9. To describe a circle that shall pass through a given point and cut equal chords of a given length from two parallels. Analysis. Let A be the given point, BC and DE the given parallels, MN the given length, and O the center of the required circle. Since the circle cuts equal chords from tw^o parallels, what must be the relative distance b of its center from each ? Therefore what line must be one locus for ? F Draw the ± bisector of MN^ cutting F(? at P. How, then, does VM compare with the radius b~M of the circle required ? How shall we then find a point O on FG that is at a distance TM from J. ? Do we then know that is the center of the required circle ? 10. To describe a circle that shall be tangent to each of two given intersecting lines. 11. To find in a given line a point that is equidistant from tAvo given points. 12. To find a point that is equidistant from two \\ "-^^ given points and at a given distance from a third p \x' q given point. 13. To describe a circle that has a given radius and passes through two given points. 14. To find a point at given distances from two given points. 15. To describe a circle that has its center in a given line and passes through two given points. 16. To find a point that is equidistant from two given points and also equidistant from two given intersecting lines. 17. To find a point that is equidistant from two given points and also equidistant from two given parallel lines. , 18. To find a point that is equidistant from c A'~~^, ^ two given intersecting lines and at a given dis- tance from a given point. 19. To find a point that lies in one side of '^ '^ a given triangle and is equidistant from the other two sides. EXEECISES 145 255. General Directions for solving Problems. In attacking a new problem draw the most general figure possible and the solution may be evident at once. If the solution is not evident, see if it depends on finding a point, in which case see if two loci can be found. If this is not the case, assume the problem solved and try to work backwards, — the method of analysis. EXERCISE 39 1. To draw a common tangent to two given circles. If the centers are and 0' and the radii r and r^, the tangent QR seems to be II to CKJlf, a tangent from 0' to a circle whose radius is r — r'. If this is true, we can easily reverse the process. Since there are two tangents from (X, so there are two common tangents. In the right-hand figure tlie tangent QR seems to be II to (XM, a tangent from CK to a circle whose radius is r + r'. If this is true, we can easily- reverse the process. There are four common tangents in general. 2. To draw a common tangent to two given circles, using the following figures. 3. The locus of the vertex of a right triangle, having a given hypotenuse as its base, is the circle described upon the given hypotenuse as a diameter. 4. The locus of the vertex of a triangle, having a given base and a given angle at the vertex, is the arc which forms with the base a segment in which the given angle may be inscribed. 146 BOOK II. PLANE GEOMETEY To construct an isosceles triangle^ having given : 5. The base and the angle at the vertex. 6. The base and the radius of the circumscribed circle. 7. The base and the radius of the inscribed circle. 8. The perimeter and the altitude. c- Let ABC be the A required, EF the given perimeter. The altitude CD passes through the ^^ -^-i middle of EF, and the A EA C, BFC are isosceles. A D B To construct a right triangle, having given : 9. The hypotenuse and one side. 10. One side and the altitude upon the hypotenuse. 11. The median and the altitude upon the hypotenuse. - 12. The hypotenuse and the altitude upon the hypotenuse. 13. The radius of the inscribed circle and one side. 14. The radius of the inscribed circle and an acute angle. To construct a triangle, having given : 15. The base, the altitude, and an angle at the base. 16. The base, the altitude, and the angle at the vertex. 1 7. One side, an adjacent angle, and the sum of the other sides. 18. To construct an equilateral triangle, hav- , — .c ing given the radius of the circumscribed circle. 19. To construct a rectangle, having given one side and the angle between the diagonals. 20. Given two perpendiculars, AB and CD, ^ intersecting in 0, and a line intersecting these perpendiculars in E and F; to con- ^ struct a square, one of whose angles shall ^- q coincide with one of the right angles at O, and the vertex of the opposite angle of the square shall lie in EF. (Two solutions.) A^ -. o -r-)iB EXERCISES 147 /« 21. A straight rod moves so that its ends con- stantly touch two fixed rods perpendicular to each other. Find the locus of its mid-point. 22. A line moves so that it remains par- allel to a given line, and so that one end lies on a given circle. Find the locus of the other end. 23. Find the locus of the mid- point of a line-segment that is drawn from a given external point to a given circle. 24. To draw lines from two given points P and Q which shall meet on a given line AB and make equal angles with AB. '.' Z BEQ = Z PEC, .'. Z CEP' = Z PEC. (Why ?) But it is easy to make Z CEP'= Z PEC, by mak- ing PP' JlAB, and CP' = PC, and joining P' and Q. 25. To find the shortest path from a point P to a line AB and thence to a point Q. q Prove that PE + EQ<PF-\- FQ, where Z BEQ = Z PEC. _i This shows that a ray of light from a point to a ^ €■ plane mirror and thence to another jwint takes the i-'-'' shortest possible path. 26. The bisectors of the angles included by the opposite sides (produced) of an inscribed quadrilateral intersect at right angles. Arc ^X— arc MD = arc XB - Arc YA - arc BN = arc DY- .'. arc YX + arc NM = arc MY + arc XN. (Why ?) .-. Z YIX = Z XIN. (Why ?) How does this prove the proposition ? Discuss the impossible case. P E^-' F B arc CM. (Why ?) arc NC. (Why ?) 148 BOOK II. PLANE GEOMETRY 27. Construct this design, making the figure twice this size. Construct the equilateral A. Then describe the small (D with half the side of the A as a radius. Then find the radius of the circumscribing O. 28. A circular window in a church has a de- sign similar to the accompanying figure. Draw it, making the figure twice this size. This is made from the figure of the preceding exer- cise, by erasing certain lines. 29. Two wheels of radii 1 ft. 6 in. and 2 ft. 3 in. respec- tively are connected by a belt, drawn straight between the points of tangency. The centers being 6 ft. apart, draw the figure mathematically. Use the scale of 1 in. to the foot. 30. A water wheel is broken and all but a fragment is lost. A workman wishes to restore the wheel. Make drawing showing how he can construct a wheel the size of the original. 31. In this figure Zm = 62°, ^ndZn=2S°. Find the number of degrees in each of the other angles, and determine whether ^^ is a diameter. 32. In this figure ZB = 4.1°, ZA = 65°, and ZBDC = 97°. Eind the number of degrees in each of the other angles, and determine whether CD is a diameter. 33. Construct or explain why it is im- ^' ~ possible to construct a triangle with sides 3 in., 2 in., 6 in. ; also one with sides 5 in., 7 in., 12 in.; also one with sides 2 in., 1 in., 1^ in. 34. Show how to draw a tangent to this circle p| at the point P, the center of the circle not being accessible. .1^^^ EXERCISES 149 EXERCISE 40 1. In a circle whose center is O the chord AB is drawn so that Z BA O = 27°. How many degrees are there in Z.AOB? 2. In a circle whose center is the chord AB is drawn so that Z.BAO = 25°. On the circle, and on the same side of AB as the center 0, the point D is taken and is joined to A and B. How many degrees are there in Z.ADB? 3. What is the locus of the mid-point of a chord of a circle formed by secants drawn from a given external point ? 4. In a circle whose center is two perpendiculars OM and ON are drawn to the chords AB and CD respectively, and it is known that Z NMO = Z ONM. Prove that AB=CD. 5. Two circles intersect at the points A and B. Through A a variable secant is drawn, cutting the circles at C and D. Prove that the angle DBC is constant. 6. Let A and B be two fixed points on a given circle, and M and N be the extremities of a rotating diameter of the same circle. Find the locus of the point of intersection of the lines AM and BN. 7. Upon a line AB a, segment of a circle containing 240° is constructed, and in the segment any chord PQ subtending an arc of 60° is drawn. Find the locus of the point of intersection of AP and BQ ; also of A Q and BP. 8. To construct a square, given the sum of the diagonal and one side. ^d Let A BCD be the square required, and CA the di- j^ ^ A/ \ agonal. Produce CA, making AE = AB. A ABC and """->.,:;, /' ABE are isosceles and ABAC = ZACB = i6°. Find B the value of Z E. Construct Z CBE. Now reverse the reasoning. The propositions in Exercise 40 are taken from recent college entrance examination papers. 150 BOOK II. PLANE GEOMETEY EXERCISE 41 Review Questions 1. Define the word circle and the principal terms used in connection with it. 2. What is meant by a central angle ? How is it measured ? 3. What is meant by an inscribed angle? How is it measured? 4. State the general proposition covering all the cases that have been considered relating to the measure of an angle formed by the intersection of two secants. 5. State all of the facts you have learned relating to equal chords of a circle. 6. State all of the facts you have learned relating to unequal chords of a circle. 7. State all of the facts you have learned relating to tangents to a circle. 8. How many points are required to determine a straight line ? two parallel lines ? an angle ? a circle ? 9. Name one kind of magnitude that you have learned to trisect, and state how you proceed to trisect this magnitude. 10. In order to construct a definite triangle, what parts must be known ? 11. What are the important methods of attacking a new problem in geometry ? Which is the best method to try first ? 12. What is meant by determinate, indeterminate, and im- possible cases in the solution of a problem ? 13. Distinguish between a constant and a variable, and give an illustration of each. 14. Distinguish between inscribed, circumscribed, and escribed circles. 15. What is meant by the statement that a central angle is measured by the intercepted arc ? BOOK III PROPORTION. SIMILAR POLYGONS 256. Proportion. An expression of equality between two equal ratios is called a proportion. 257. Symbols. A proportion is written in one of the fol- lowing forms : 7 = ~;5 a:b = c : d\ a : b : : c : d. This proportion is read " a is to 6 as c is to d " ; or " the ratio of a to 5 is equal to the ratio of c to d." 258. Terms. In a proportion the four quantities compared are called the terms. The first and third terms are called the antecedents; the second and fourth terms, the consequents. The first and fourth terms are called the extremes; the second and third terms, the means. Thus in the proportion a:b = c :d, a and c are the antecedents, b and d the consequents, a and d the extremes, b and c the means. 259. Fourth Proportional. The fourth term of a proportion is called the fourth proportional to the terms taken in order. Thus in the proportion a : 6 = c : d, d is the fourth proportional to a, &, and c. 260. Continued Proportion. The quantities a, b, c, d, • • • are said to be in continued proportion, if a : b = b : c = c : d = • - • . If three quantities are in continued proportion, the second is called the mean proportional between the other two, and the third is called the third proportional to the other two. Thus in the proportion a-.b = b:c,b is the mean proportional between a and c, and c is the third proportional to a and b. 151 152 BOOK III. PLANE GEOMETRY Proposition I. Theorem 261. In any proportion the product of the extremes is equal to the product of the means. Given a:b = c:d. To prove that ad = he. Proof. j = ^' §257 a Multiplying by bd, ad = bc, by Ax. 3. " q.e.d. 262. Corollary 1. The mean proportional between tivo quantities is equal to the square root of their product. For if a:h = h:c, then h^ = ac (§ 261), and h = Vac, by Ax. 5. 263. Corollary 2. If the two antecedents of a proportion are equals the two consequents are equal. 264. Corollary 3. If the product of two quantities is equal to the product of two others^ either two may be made the extremes of a proportion in which the other two are made the means. For if ad = 6c, then, by dividing by 6d, - = -, by Ax. 4. 6 d Proposition II. Theorem 265. If four quantities are in proportion, they are in proportion hy alteimation ; that is, the first term is to the third as the second term is to the fourth. or Given a: b = c:d. To prove that a: c = b: d. Proof. ad = be. §261 Dividing by cd, a b -c = d' Ax. 4 a:c = b:d,hy % 257. Q.E.D. THEORY OF PROPORTION 153 Proposition III. Theorem 266. If four quantities are in proportion, they are in proportion hy inversion ; that is, the second term is to the first as the fourth term is to the third. Given a:b = c:d. To prove that h: a = d: c. Proof. bc = ad. §261 Dividing each member of the equation by ac, - = -> Ax. 4 a c or b : a = d : c, hy ^ 257. Q. e. d. Proposition IV. Theorem 267. If four quantities are in proportion, they are in proportion hy composition; that is, the sum of the first two terms is to the second term as the sum of the last tivo terms is to the fourth term. Given a:b=c:d. To prove that a-\-h :h = c -\- d: d. Proof. j = -- §257 d Adding 1 to each member of the equation, a ^ G ^ a-\-b c-\-d b ^ d ' .-. a-[-b:b = c + d:d,hy %257. Q.e.d. In a similar manner it may be shown that a -{- b : a = c -\- d : c. a ^ c ^ . ^ - + 1 = -+1, Ax.l or 154 BOOK III. PLANE GEOMETRY Proposition V. Theorem 268. If four quantities are in proportion , they are in proportion hy division; that is, the difference of the first two terms is to the second term as the difference of the last two terms is to the fourth term. §257 Ax. 2 or Given a:b = c:d. To prove that a — b: h = c — d: d. Proof. a c h~d' "h ^ d ^' a — h c — d h d .'.a-h:h = c-d:d,hj §257. In a similar manner it may be shown that a — b:a = c — d:c. Q.E.D. Proposition VI. Theorem 269. Li a series of equal ratios, the sum of the ante- cedents is to the su7n of the consequents as any ante- cedent is to its consequent. Given a:b = c:d=:e:f=g:h. To prove that a-\- e -\- e-^ g:h -\- d +/+ h = a:h. Proof. Let r = - = - = - = 'z-- d J I I Then a = hr, c = dr, e=fr, g = hr. Ax. 3 ,',a.\-e + e-\-g = {h + d+f+h)r. Ax. 1 ^ CL -\- c -\- e -\- g a Ax. 4 "b-\-d+f+h h a^c-\-e + g:b-\-d +/+ h = a : h, by § 257. Q.E.D. THEORY OF PROPORTION 155 Proposition VII. Theorem 270. Like powers of the terms of a proportion are iri proportion. Given a:bz=c: d. To prove that a" : 5" = ^ : (^. Proof. X = ^- ' §257 a •*• tt^'t:' by Ax. 5. q.e.d. Proposition VIII. Theorem 271. If three quantities are in continued proportion, the first is to the third as the square of the first is to the square of the second. Given a:b=b:c. To prove that a: c = a^:b\ Proof. a^ = a% Iden. and ae = b\ a" a a^ ac c b^ §261 Ax. 4 . • . a :c = a'^: h% by § 257. Q.E.D. 272. Nature of the Quantities in a Proportion. Although we may have ratios of lines, or of areas, or of solids, or of angles, we treat all of the terms of a proportion as numbers. If b and d are lines or solids, for example, we cannot multiply each member of - = - by bd, as in § 261. Hence when we speak of the product of two geometric viagni- tudes, we mean the product of the numbers that represent them when expressed in terms of a common unit. 166. BOOK III. PLANE GEOMETEY EXERCISE 42 1. Prove that a:h = ma : mfib. 2. \i. a\h=^G\d^ and m : 71=.]) : q, prove that am :bn=cp : dq. If a:b = c: d, prove the following : 3. a:d = bc:d\ 7. ma : nb = me : nd. 4. l:b = c: ad. S. a — l:b = bc — d:bd. 5. ad:b = c:l. 9. a-^l:l = bc + d:d. 6. ma : b = mc : d. 10. 1 : ^c = 1 : ad. 11. a-\-b:a — b = c-\-d:c — d. In Ex. 11, use § 267 and § 268, and Ax. 4. In this case a, 6, c, and d are said to be in proportion by composition and division. If a:b = b: c, prove the following : 12. G'.b = b'.a. 14. (^ + V^) (Z> - V^) = 0. 13. a:c = P:c''. 15. ac -l:b -1 = b -\-l'.l. 16. If 2:7= 3: a-, show that 2^^ = 21, andic=10^. ^mc? ^Ae value of x in the following : 17. l:7=3:ic. 29. ic : 2.7 = 7: 5.4. 18. 2:9 = 5:ic. 30. cc : 8.1 = 0.3 : 0.9. 19. 4:28 = 3:ic. 31. 2:ic = £c:32. 20. 2:^ = a;:12. 32. 7:£c = cc:28. 21. 3:5 = ;:c:9. 33. l:l-f-x = a; -1: 3. 22. 7:21=£c:5. 34. 5 :£c - 2 = a: + 2 :1. 23. 3:5 = £c 4-1:10. 35. cc2:2a = 3a:6. 24. 8:15 = 2ic + 3:45. 36. x'.4.a = 2a^:x\ 25. 0.8:ic = 4:9. 37. a:l = x-l:l. 26. 0.7:a; = 21:15. 38. a; +l:ic -1= 3 : 2. 27. 0.25:cc = 5:8. 39. 3 :cc + 4 = a; - 4: 3. 28. a;:1.3 = 4:0.26. 40. ab:b = b-cx'.bc-x. PKOPORTIOKAL LINES 157 Proposition IX. Theorem 273. If a line is drawn through two sides of a tri- angle parallel to the third side, it divides the tivo sides proportionally. Given the triangle ABC, with EF drawn parallel to BC. To prove that EB:AE=FC: AF. Case 1. When AE and EB are commensurable. Proof. Assume that MB is a common measure oi AE and EB. Let MB be contained m times in EB, and n times in AE. Then EB:AE = m:n. {For m and n are the numerical measures of EB and AE.) At the points of division on EB and AE draw lines II to BC. These lines will divide A C into m-\-n equal parts, of which FC will contain m parts, and AF will contain n parts. § 134 .•.FC:AF=m:n. .'.EB:AE = FC:AF,hyAx.S. Q.e.d. For practical purposes this proves the proposition, for even if AE and EB are incommensurable, we can, by taking a unit of measure small enough, find the measure of AE and EB to as close a degree of approxi- mation as we may desire, just as we can carry V2 to as many decimal places as we wish, although its exact value cannot be expressed rationally. On this account many teachers omit the incommensurable case dis- cussed on page 158, or merely require the proof there given to be read aloud and explained by the class. 158 BOOK III. PLANE GEOMETEY Case 2. When AE and EB are incommensurable, A B (J Proof. Divide AE into a number of equal parts, and apply one of these parts to EB as many times as EB will contain it. Since AE and EB are incommensurable, a certain number of these parts will extend from E to some point G', leaving a remainder GB less than one of these parts. Draw GH II to BC. Then EG:AE = FN : AF. Case 1 By increasing the number of equal parts into which AE is divided, we can make the length of each part less than any assigned positive value, however small, but not zero. Hence GB, which is less than one of these equal parts, has zero for a limit. § 204 And the corresponding segment HC has zero for a limit. Therefore EG approaches EB as a limit, and FH approaches EC as a limit. .*. the variable — — approaches -~— as a limit, AJii All/ and the variable — — approaches — — as a limit. AF ^^ AF But — — is always equal to — — • Case 1 .•.f| = f|,by.207. A Fl L \« \ \m I \\ PKOPORTIONAL LINES 159 274. Corollary 1. One side of a triangle is to either of its segments cut off hy a line parallel to the base as the third side is to its corresponding segment. For EB.AE = FC.AF. §273 By composition, EB + AE: AE = FC + AF: AF, § 267 or AB:AE = AC:AF. Ax. 11 275. Corollary 2. Three or more parallel lines cut off proportional intercepts on any two transversals. Draw ^iV II to CD. Then AL = CG, LM = GK, MN = KB. § 127 Now AH:AM= AF:AL=FH:LM, §274 and AH:AM=HB:MN. §273 .'.AF'.CG =FH:GK=HB:KB. Ax.9 b N D EXERCISE 43 1. In the figure of § 275, suppose AH = 5 in., AF=2 in., and CK=6 in. Find the length of CG. d c 2. In this square PQ is II to AB. If a side of the p square is 10 in., Z)^ = 14.14 in. If DP = 3 in., what is the length ot BQ? 3. The sides of a triangle are respectively 3 in., 4 in., and 5 in. A line is drawn parallel to the 4-inch side, cutting the 3-inch side 1 in. from the vertex of the largest angle. Find the length of the two segments cut from the longest side. 4. Two pieces of timber 1 ft. wide are fitted together at right angles as here shown. AB is 8 ft. long, AC 6 ft. long, and the distance BCj along the dotted line, is 10 ft. A carpenter finds it necessary to saw along the dotted line. Find the length of the slanting cut across the upright piece ; across the horizontal piece. 160 BOOK III. PLANE GEOMETEY Proposition X. Theorem 276. If a line divides two sides of a triangle jpro^or- tionally, it is parallel to the third side. or B G Given the triangle ABC with EF drawn so that EBFC AE~ AF' To prove that EF is II to BC. Proof. Suppose that EF is not parallel to BC. Then from E draw some other line, as EH, parallel Then AB : AE = AC : AH. {One side of a A is to either of its segments cut off by a line II base as the third side is to its corresponding segment.) But EB:AE = FC:AF. .-. EB + AE: AE = EC -\- AFiAF, AB:AE = AC:AF. .'.AC:AF=AC:AH. .\AF=AH. {For the two antecedents are equal.) .'. EF and EH must coincide. {For their end points coincide.) EH is II to BC. .'. EF, which coincides with EH, is II to BC. This proposition is the converse of Prop. IX. But to^C. §274 to the Given §267 Ax. 11 Ax. 8 §263 Post. 1 Const. Q.E.D. PKOPOETIONAL LINES 161 277. Dividing a Line into Segments. If a given line AB is divided at P, a point between the extremities A and B, it is said to be divided internally into the segments AP and PB'j and if it is divided at P', a point in the prolongation of BA, it is said to be divided externally into the segments AP' and P'B. P' A P B In either case the length of the segment is the distance from the point of division to an extremity of the line. If the line is divided internally, the sum of the segments is equal to the line ; and if the line is divided externally, the difference of the segments is equal to the line. Suppose it is required to divide the given line AB internally and externally in the same ratio ; as, for example, in the ratio of the two numbers 3 and 5. P' APE We divide AB into 3 + 5, or 8, equal parts, and take 3 parts from A ; we then have the point P, such that AP'.PB = Z:n. (1) Secondly, we divide AB into 5 — 3, or 2, equal parts, and lay off on the prolongation of BA three of these equal parts ; we then have the point P', such that ylP':P'5 = 3:5. (2) Comparing (1) and (2), we have AP:PB = AP''.P'B. 278. Harmonic Division. If a given straight line is divided internally and externally into segments having the same ratio, the line is said to be divided harmonically. Thus the line AB has just been divided internally and externally in the same ratio, .3 : 5, and ^ JB is therefore said to be divided harmonically at P and P' in the ratio 3:5. 162 BOOK III. PLANE GEOMETRY Proposition XI. Theorem 279. The bisector of an angle of a triangle divides the opposite side into segmeiits which are proportional to the adjacent sides. '--^G Given the bisector of the angle C of the triangle ABC^ meeting ABatM, To prove that AM: MB = CA: CB. Proof. Erom A draw a line II to MC. This line must meet BC produced, because BC and MC cannot both be parallel to the same line. § 94 Let this line meet BC produced at E. Then AM:MB=EC :CB. §273 {If a line is drawn through two sides of a A parallel to the third side, it divides the two sides proportionally.) Also ZACM=ZCAE, § 100 {Alt.-int. A of II lines are equal.) and Z MCB = ZAEC. § 102 {Ext.-int. A of II lines are equal.) But ZACM=ZMCB. Given ,'.ZCAE = ZAEC. Ax. 8 .\EC=CA. §76 Put CA for its equal ^C in the first proportion. Then AM:MB= CA :CB,hy Ax. 9. Q. E. D. PEOPOETIONAL LINES 163 Pboposition XII. Theorem 280. The bisector of an exterior angle of a triangle divides the opposite side externally into segments ivhich are proportional to the adjacent sides. A ^ Given the bisector of the exterior angle ECA of the triangle ABCy meeting BA produced at M'. To prove that AM' : M'B = CA:CB. Proof. Draw AF II to M'C, meeting EC at F. Then AM': M'B=FC -.CB. §274 {One side of a A is to either of its segments cut off by a line II to the base as the third side is to its corresponding segment.) Now Z ECM'= Z CFA, § 102 and ZM'CA=ZFAC. §100 But ZECM'=ZM'CA. Given .'.ZCFA=ZFAC. Ax. 8 .-. CA = FC. § 76 Put CA for its equal FC in the first proportion. Then AM': M'B = CA : CB, by Ax. 9. Q.e.d. Discussion. In case CA = CB, what is the arrangement of the lines ? 281. Corollary. The bisectors of the interior angle arid the exterior angle at the same vertex of a triangle, meeting the opposite side, divide that side harmonically. 164 BOOK III. PLANE GEOMETRY EXERCISE 44 1. In a triangle ABC, AB = 6.5, CA = 6,BC = 7. Find the segments oi AB made by the bisector of the angle C 2. In a triangle ABC, CA = 7.5, BC = 7,AB = 8. Find the segments of CA made by the bisector of the angle B. 3. The sides of a triangle are 12, 16, 20. Find the segments of the sides made by bisecting the angles. 4. If a spider, in making its web, makes A'B' II to AB, B'C II to BC, C'D' II to CD, D'E' II to DE, and E'F' II to EF, and then runs a line from F' II to FA, will it strike the point A ' ? Prove it. 5. From any point O within the triangle ABC the lines OA, OB, OC are drawn and are bisected respectively by A', B', and C. Prove that Z CBA = Z C'B'A'. 6. Prove Ex. 5 if the point is outside the triangle. 7. From any point within the quadrilateral ABCD lines are drawn to the vertices A,B,C, D, and are bisected by A',B',C', D'. Prove that Z CBA = Z C'B'A '. 8. If a pendulum swings at the point 0, cutting two paral- lel lines at P and Q respectively, the ratio OP : OQ is constant. 9. Through a fixed point P a line is drawn cutting a fixed line at X. PX is then divided at Y so that the ratio PYiYX is constant. Find the locus of the point F as Z moves along the fixed line. 10. From the point P on the side CA of the triangle ABC parallels to the other sides are drawn meeting ABin Q and BC in R. Prove that AQ: QB = BR : RC. 11. In the triangle ABC, P and Q are taken on the sides CA and BC so that AP : PC = BQ: QC. AR is then drawn parallel to PB, meeting CB produced in R. Prove that CB is the mean proportional between CQ and CR. SIMILAR POLYGONS 165 282. Similar Polygons. Polygons that have their correspond- ing angles equal, and their corresponding sides proportional, are called similar polygons. Thus the polygons ABODE and A'B'C'iyE' are similar, if the A A, B, C, D, E are equal respectively to the A A% B% C", i/, E\ and if AB : A'B' = BC : B'C = CD : C'XK = DE : D'E' = EA : E'A\ Similar polygons are commonly said to be of the same shape. 283. Corresponding Lines. In similar polygons those lines that are similarly situated with respect to the equal angles are called corresponding lines. Corresponding lines are also called homologous lines. 284. Ratio of Similitude. The ratio of any two corresponding lines in similar polygons is called the ratio of similitude of the polygons. The primary idea of similarity is likeness of form. The two conditions necessary for similarity are : 1. For every angle in one of the figures there must he an equal angle in the other. 2. The corresponding sides must he proportional. Thus Q and Q! are not similar ; the corresponding sides are proportional, but the corresponding angles are not equal. Also R and it' are not similar; the corresponding angles are equal, but the corresponding sides are not proportional. Q' In the case of triangles either condition implies the other. 166 BOOK III. PLANE GEOMETRY Proposition XIII. Theorem 285. Two rnutually equiangular triangles are similar. Given the triangles ABC and A'B^C\ having the angles A, B^ C equal to the angles A', B\ C respectively. To prove that the A ABC and A'B'C' are similar.. Proof. Since the A are mutually equiangular, Given we have only to prove that AB : A^B' = AC : A'C = BC : B'C. § 282 Place the A^'^'C*' on the A^^C so thatZC shall coincide with its equal, the Z C, and A'B' take the position PQ. Post. 5 Then Aj^ = ^ A. Given .-. PQ is II to AB. § 103 .'.AC:PC = BC: QC ; §274 that is, AC :A'C' = BC iB'C. Ax. 9 Similarly, by placing the A A'B'C on the A ABC so that Z B' shall coincide with its equal, the Z B, we can prove that AB:A'B' = BC:B'C'. .\ AB : A'B' = AC : A'C = BC : B'C. Ax. 8 .-. A ABC is similar to A A'B'C, by § 282. q.e.d. 286. Corollary 1. Two triangles are similar if two angles of the one are equal respectively to two angles of the other. 287. Corollary 2. Two right triangles are similar if an acute angle of the one is equal to an acute angle of the other. SIMILAR POLYGONS 167 Proposition XIV. Theorem 288. If two triangles have an angle of the one equal to an angle of the other ^ and the including sides propor- tional, they are similar. Given the triangles ABC and A'B'C\ with the angle C equal to the angle C and with CA : C'A' = CB : C^B\ To prove that the A ABC a7id A'B'C are similar. Proof. Place the A A'B'C on the A ABC so that Z.C' shall coincide with its equal, the Z C. Post. 5 Then the A A'B'C will take the position of the A PQC. CA _ CB , CA'~C^'' CA CB CP~ CQ CA-CP CB — CQ Now that is, Given Ax. 9 or CP PA CP ..'. PQ is CQ QB CQ to AB. {If a line divides two sides of a A proportionally, it is .'. Ap =z Z.A, and Aq = Z.B. Now /LC = AC'. .'. A PQC is similar to A ABC. .'. A A'B'C is similar to A ABC. 268 §276 to the third side.) §102 Given §285 Q. E. D. 168 BOOK III. PLANE GEOMETRY Proposition XY. Theorem 289. If two triangles have their sides respectively proportional^ they are similar. A B A' B' Given the triangles ABC and A'B'C\ having AB : A'B' =BC:B'C'=CA: CA \ To prove that the A ABC and A'B'C are similar. Proof. Upon CA take CP equal to CA\ and upon CB take CQ equal to C'^'; and draw PQ. Now Or, since Also CA : C'A' = BC:B'C'. CP = CA', and CQ = C'B', CA: CP = CB:CQ. ZC = ZC. A ABC and,PQC are similar. Given Const. Ax. 9 Iden. §288 {If two A have an angle of the one equal to an angle of the other, and the including sides proportional, they are similar.) that is, But But .\CA:CP = AB:PQ; CA : CA' = AB : PQ. CA:CA' = AB:A'B'. .-. AB:PQ = AB:A'B'. .\PQ = A'B'. Hence the A PQC and A'B'C Sive congruent. APQC has been proved similar to A ABC. .'.A A'B'C is similar to A ABC §282 Ax. 9 Given Ax. 8 §263 §80 Q.E.D. SIMILAR POLYGOJ^S Proposition XVI. Theorem 169 290. TiDo triangles ivhich have their sides respectively jjarallel, or ^respectively perpendicular, are similar. Given the triangles ABC and A'B'C'j with their sides respec- tively parallel ; and the triangles DEF and D'E'F'y with their sides respectively perpendicular. To prove that 1. the A ABC and A'B'C are similar ; 2. the A DEF and D'E'F' are similar. Proof. 1. Produce EC and AC to B'A', forming A x and y. Then Z.B = Zx(^ 100), Siud ZB' = Zx. §102 .\ZB = ZB'. Ax. 8 In like manner, ZA = ZA'. .-. A ABC is similar to A A'B'C. § 286 2. Produce DE and FD to meet D'E' and F'D' at P and R. The quadrilateral E'QEP has zip and q right angles. Given .*. A E' and PEQ are supplementary. § 144 But A y and Pi?Q are supplementary. § 43 Therefore Zy = ZE'. § 58 In like manner, Zx = ZD'. .-.A X>£:f is similar to A D'E'F', by § 286. Q.e.d. Discussion. The parallel sides and the perpendicular sides respectively are corresponding sides of the triangles. 170 BOOK III. PLANE GEOMETRY Pkoposition XVII. Theorem 291. The perimeters of two similar polygons have the same 7'atio as any two corresponding sides. D Given the two similar polygons ABCDE and A'B'C'D'E', with p and />' representing their respective perimeters. To prove that p ip' = AB : A'B'. AB _ BC_ _ CD_ _ DE _ EA_ A'B' ~ B'C ~ CD' ~ D'E' ~ EA'' ^ AB -{- BC -{- CD -\- DE -{- EA _ AB A 'B' + B'C + C"X>'+ D'E' + E'A'~ A 'B'' §269 r,2):p' = AB:A'B'jhyAx.9. q.e.d. EXERCISE 45 1. The corresponding altitudes of two similar triangles have the same ratio as any two corresponding sides. 2. The base and altitude of a triangle are 15 in. and 7 in. respectively. The corresponding base of a similar triangle is 3.75 in. Find the corresponding altitude. 3. If two parallels are cut by three concurrent transversals, the corresponding segments of the parallels are proportional. 4. The point P is any point on the side OX of the angle XOY. From P a perpendicular PQ is let fall on OY. Prove that for any position of P on OX the ratio OP : PQ is constant, and the ratio PQ: OQ is constant. SIMILAR POLYGONS 171 5. In drawing a map of a triangular field with sides 75 rd., 60 rd., and 50 rd. respectively, the longest side is drawn 1 in. long. How long are the other two sides drawn ? 6. This figure represents part of a diagonal scale used by draftsmen. The distance from to 10 is 1 centimeter, or 10 millimeters. Show how to measure 5 mm, ; 1 mm. ; 0.9 mm. ; 0.5 mm. ; 1.5 mm. On what proposition does this depend ? _. _ . 7. This figure represents a pair of proportional compasses used by draftsmen. By adjusting the screw at 0, the lengths OA and OC, and the corre- sponding lengths OB and OD, may be varied pro- portionally. Prove that AOAB is always similar to AOCD. If 0^ = 3 in. and OC = 5 in., then AB is what part of CD? 8. ABCD is any polygon and P is any point. On ^ P any point A ' is taken and A 'B' is drawn parallel to AB as shown. Then B'C'&nd CD' are drawn parallel to BC and CD. Is D'A' parallel to DA ? Is A'B'C'D' similar to ABCD? Prove it. 9. If two circles are tangent externally, the corresponding segments of two lines drawn through the point of contact and terminated by the circles are proportional. 10. If two circles are tangent externally, their common ex- ternal tangent is the mean proportional between their diameters. 11. AB and AC are chords drawn from any point /I on a circle, and AD is ^ diameter. If the tangent at D intersects AB and A C at i? and F, the triangles ABC and A EF are similar. 12.' If AD and BE are two altitudes of the triangle ABC^ the triangles DEC and ABC are similar. p^-: 172 BOOK III. PLANE GEOMETRY Proposition XVIII. Theorem 292. If tivo polygons are similar, they can he sepa- rated hito the same number of triangles, similar each to each, and similarly placed. Given two similar polygons ABCDE and A'B'CD'E^ with angles A, B, C, 2>, E equal to angles ^', 5', C, D\ E' respectively. To prove that ABODE and A'B'C'D'E' can be separated into the same number of triangles, similar each to each, and similarly placed. Proof. Draw the corresponding diagonals DA, D'A', and DB, D'B'. Since Z.E = Z.E\ §282 and DE:D'E' = EA : E'A', §282 .'. ADEA and D'E'A' are similar. §288 In like manner, ADBC and D'B'C are similar. Purthermore ZBAE = ZB'A'E', §282 and ZDAE = ZDA'E'. §282 By subtracting, ZBAD = ZBA'D'. Ax. 2 Now DA :DA' = EA : E'A', §282 and AB:A'B' = EA : E'A'. §282 . '. DA:D'A' = AB:A'B'. Ax. 8 .-.A DAB and D'A'B' are similar. by § 288. Q.E.D. SIMILAR POLYGONS 173 Proposition XIX. Theorem 293. If tivo polygons are composed of the same num- ber of triangles, similar each to each, and similarly pjlaced, the polygons are similar. A B A' B' Given two polygons ABCDE and A'B^C'D'E' composed of the triangles DEA^ DAB^ DBC, similar respectively to the triangles D'E'A\ D'A'B', D'B'C, and similarly placed. To prove that ABCDE is mnilar to A'B'C'D'U'. Proof. ZE = ZE'. Also ZDAE = Z D'A 'E'j and ZBAD = ZB'A'D'. By adding, ZBAE = ZB'A 'E'. Similarly Z CBA = Z C'B'A ', and Z EDC = Z E'D'C. Again, ZC = ZC'. § 282 Hence the polygons are mutually equiangular. DE _ EA DA AB DB BC CD ~ E'A' §282 §282 Ax. 1 Also 282 D'E' E'A' D'A' A'B' D'B' B'C CD' Hence the polygons are not only mutually equiangular but they have their corresponding sides proportional. Therefore the polygons are similar, by § 282. q.e.d. This proposition is the converse of Prop. XVIII. 174 BOOK III. PLANE GEOMETRY Proposition XX. Theorem 294. If m a right triangle a perpeiidicular is clraivn from the vertex of the right aiigle to the hypotenuse : 1. The triangles thus formed are similar to the given triangle, and are similar to each other. 2. The perpendicidar is the mean proportional he- tween the segments of the hypotenuse. 3. Each of the other sides is the mean proportional hetiveen the hypotenuse and the segment of the hypote- nuse adjacent to that side, c A F B Given the right triangle ABC^ with CF drawn from the vertex of the right angle C, perpendicular to AB. 1. To prove that the A BCA, CFA, BFC are similar. Proof. Since the Z. a^ is common to the rt. A CFA and BCA, .'. these A are similar. §287 Since the Z Z> is common to the rt. A BFC and BCA, .'. these A are similar. § 287 Since the A CFA and BFC are each similar to A BCA, .'. these A are mutually equiangular, § 282 Therefore the A CFA and BFC are similar, by § 285. Q. e. d. 2. To prove that AF :CF=CF: FB. Proof. In the similar A CFA and BFC, AF: CF=CF:FB, by § 282. q.e.d. NUMEKICAL PROPERTIES OF FIGURES 175 3. To prove that AB:AC = AC:AF, and AB:BC = BC:BF. Proof. In the similar A EC A and CFA, AB:AC = AC:AF. §282 In the similar ABC A and BFC, AB:BC = BC: BF, by § 282. q.e.d. 295. Corollary 1. The squares on the two sides of a right triangle are proportional to the segments of the hypotenuse adjacent to those sides. , From the proportions in § 294, 3, AAJ^ = ABx AF, and liC^ = AB x BF. § 261 Hence -^^ AB^AF ^A^ ^^ , BC^ AB X BF BF 296. Corollary 2. The square on the hypotenuse and the squai^e on either side of a right triangle are proportional to the hypotenuse and the seg7nent of the hypotenuse adjacent to that side. Since AB^ ^ AB x AB, Iden. and, as in § 295, AC^ = AB x AF, § 261 AB^ _ ABx AB _AB ^^ ^ '' AC''~ ABx AF~AF' 297. Corollary 3. The perpendicular from any point on a circle to a diameter is the mean proportional hehveen the segments of the diameter. g^ 298. Corollary A. If a perpendicular is drawn from any point on a circle to a diameter^ the chord from that point to ^ either extremity of the diameter is the mean proportional he- tiveen the diameter and the segment adjacent to that chord. 176 BOOK III. PLANE GEOMETRY EXERCISE 46 1. The perimeters of two similar polygons are 18 in. and 14 in. If a side of the first is 3 in., find the corresponding side of the second. 2. In two similar triangles, ABC and A'B'C', AB = 6 in., BC = 7 in., CA = S in., and A'B' = 9 in. Eind B'C and C'A'. 3. The corresponding bases of two similar triangles are 11 in. and 13 in. The altitude of the first is 6 in. Find the corresponding altitude of the second. 4. The perimeter of an equilateral triangle is 51 in. Eind the side of an equilateral triangle of half the altitude. 5. The sides of a polygon are 2 in., 2^ in., 3| in., 3 in., and 5 in. Eind the perimeter of a similar polygon whose longest side is 7 in. 6. The perimeter of an isosceles triangle is 13, and the ratio of one of the equal sides to the base is 1§. Eind the three sides. 7. The perimeter of a rectangle is 48 in., and the ratio of two of the sides is f . Eind the sides. 8. In drawing a map to the scale y^^oVoo) what length will represent the sides of a county that is a rectangle 25 mi. long and 10 mi. wide ? Answer to the nearest tenth of an inch. 9. Two circles touch at P. Through P three lines are drawn, meeting one circle in A, B, C, and the other in A\ B', C respectively. Prove the triangles ABC, A'B'C' similar. 10. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided pro- portionally by the smaller circle. 11. In an inscribed quadrilateral the product of the diagonals is equal to the sum of the products of the opposite sides. Draw D^, making ZEDC = Z ABB. The A ABB and ECB are similar ; and the A BCB and AEB are similar. NUMERICAL PROPERTIES OF FIGURES 177 Proposition XXI. Theorem 299. If tivo chords intersect ivithin a circle, the prod- uct of the segments of the one is equal to the product of the segments of the other. Given the chords AB and CDy intersecting at P. To prove that PAxFB=FCx PD. Proof. Draw A C and BD. Then since Aa = Z.a\ § 214 (Each is measured by | arc CB.) and Zc = Zc\ §214 {Each is measured by | arc DA.) .-. the A CPA and BPD are similar. § 286 .-. PA :PD=PC:PB. §282 .-. PAxPB=PC xPD,hj ^261. Q.E.D. 300. Corollary. If two chords intersect within a circle, the segments of the one are reciprocally proportional to the segments of the other. This means, for example, that PA : PD equals the reciprocal of PB : PC, or equals PC : PB, as shown above. 301. Secant to a Circle. A secant from an external j^oint to a circle is understood to mean the segment of the secant that lies between the given external point and the second j^oint of inter- section of the secant and circle. 178 BOOK III. PLANE GEOMETRY Proposition XXII. Theorem 302. If from a point outside a circle a secant and a tangent are draion, the tangent is the 7nean propor- tional hetiveen the secant and its external segment. Given a tangent AD and a secant AC drawn from the point A to the circle BCD. To prove that AC. AD = AD: AB. Proof. Draw DC and DB. Now Z c is measured by ^ arc DB, § 214 and Z c' is measured by i arc DB. § 220 .'.Z.c = Z.c\ Then in the A ADC and ABD, Z.a = /.a, Iden. and Ac — Ac\ .-. A^Z>C and ^5/)are similar. §286 .•.AC\AD=.AD'.AB, by §282. q.e.d. 303. Corollary. If from a fixed 'point outside a circle a secant is dratvn, the product of the secant and its external segment is constant in whatever direction the secant is drawn. Smce AC:AD = AD:AB, §302 .-. AC X AB = AD^ §261 Since AD is constant (§ 192), therefore AC x AB is constant. NUMERICAL PROPERTIES OF FIGURES 179 Proposition XXIII. Theorem 304. The square on the bisector of an angle of a tri- angle is equal to the product of the sides of this angle diminished hy the product of the segments made by the bisector upon the third side of the triangle. ~--C D Given the line CP bisecting the angle ACB of the triangle ABC. To prove that CP'' = CA x BC-AP x PB. Proof. Circumscribe the OBCA about the A ABC. § 240 Produce CP to meet the circle in Z>, and draw BD. Then in the A BCD and PC A, and /.m = Z. m\ Given Aa' = A a. §214 {Each is measured by I arc BC.) •. the A BCD and PC A are similar. §286 .-. CD:CA=BC: CP. §282 .-. CAxBC=CDX CP §261 = {CP-\-PD)CP Ax. 9 = CP'' + CP X PD. CPXPD=APXPB. §299 .-. CA xBC=CP^-{-APxPB. Ax. 9 CP^ =CAxBC -APx PB, by Ax. 2. Q.E.D. But This theorem enables us to compute the bisectors of the angles of a triangle terminated by the opposite sides, if the sides are known. The theorem may be omitted without destroying the sequence. 180 BOOK III. PLANE GEOMETRY Proposition XXIV. Theorem 305. In any triangle tlie jproduct of two sides is equal to the product of the diameter of the circumscribed circle hy the altitude upon the third side. Given the triangle ABC with CP the altitude, ADBC the circle circumscribed about the triangle ABC^ and CD a diameter. To prove that CAxBC=CDx CP, Proof. Draw BD. Then in the A APC and DBC, Z CPA is a rt. Z, Given Z CBD is a rt. Z, § 215 Z <t is measured by \ arc BC, and Z a' is measured by \ arc BC. § 214 .\Aa = Z.a\ .-. A APC and D^C are similar. § 287 ( Two rt. A are similar if an acute Z of the one is equal to an acute Z of the other.) .-. CA : CD=CP:BC. § 282 .'. CA XBC=CDX CP, by § 261. q.e.d. This theorem may be omitted without destroying the sequence. Props. XXIII and XXIV are occasionally demanded in college entrance ex- aminations, but they are not necessary for proving subsequent propo- sitions or for any of the exercises. Teachers may therefore use their judgment as to including them. NUMERICAL PROPERTIES OF FIGURES 181 EXERCISE 47 1. The tangents to two intersecting circles, drawn from any point in their common chord produced, are equal. 2. The common chord of two intersecting circles, if produced, bisects their common tangents. 3. If two circles are tangent externally, the common internal tangent bisects the two common external tangents. 4. If a line drawn from a vertex of a triangle divides the opposite side into segments proportional to the adjacent sides, the line bisects the angle at the vertex. 5. If three circles intersect one another, the common chords are concurrent. Let two of the chords, AB and CD, meet at 0. Join the point of intersection E to 0, and suppose that EO produced meets the same two circles at two different points P and Q. Then prove that OP = OQ {^ 299), and hence that the points P and Q coincide. 6. The square on the bisector of an exterior angle of a triangle is equal to the product of the segments determined by this bisector upon the opposite side, diminished by ^ the product of the other two sides. / '-/-^o^ Let CD bisect the exterior ZBCH of the A ABC. ^^ A ADC and FBC are similar (§ 286). Apply § 303. 7. If the line of centers of two circles meets the circles at the consecutive points A, B, C, D, and meets the common external tangent at P, then PA xPD = PBx PC. 8. The line of centers of two circles meets the common external tangent at P, and a secant is drawn from P, cutting the circles at the consecutive points E, F, G, II. Prove that PE XPH=PFXPG. Draw radii to the points of contact, and to E, P, G, H. Let fall Js on PR from the centers of the ©. The various pairs of A are similar. 182 BOOK III. PLANE GEOMETEY Proposition XXV. Problem 306. To divide a given line into parts proportional to any number of given lines. A M' n' B ^^ ^ V-^N \^^ \ \ \ \ "V \ \ n-^\ \ "^ ^^\\ n p-\ p P-x Given the lines AB^ m, n, and p. Required to divide AB into parts proportional to m, n, and p. Construction. Draw AX, making any convenient Z. with AB. On AX take AM equal to m, MN equal to n, and NP equal to j9. Draw BP. From N draw NN' II to PB, and from M draw MM' II to PB. Then M' and iV' are the division points required. q.e.f. Proof. Through A draw a line II to PB. AM' _ M'N' N'B AM ~ MN ~ NP' {Three or more || lines cut off proportional intercepts on any two transversals.) Substituting 7«,, n, and^^^ for their equals AM, MN, and NP, AM' M'N' N'B , ^ we have = = Ax. 9 7)1 n p This means that AB has been divided as required. q.e.d. In like manner, we may divide AB into parts proportional to any number of given lines. PROBLEMS OF CONSTRUCTION 183 Proposition XXVI. Phoblem 307. To find the fourth proportional to three given lines. A____yii_ B n_ G__^ \r Given the three lines m, n, and p. Required to find the fourth proportional to /«, w, and p. Construction. Draw two lines ^A' and AY containing any convenient angle. On AX take AB equal to m, and take BC equal to n. On ^ F take AD equal to p. Draw BD. From C draw CE II to BD, meeting A Y at E. Then DE is the fourth proportional required. q.e.f. Proof. AB : BC = AD : DE. § 273 (If a line is drawn through tioo sides o/a A II to the third side, it divides the two sides proportionally.) Substituting 7ti, n, and^ for their equals AB, BC, and AD, we have m:n=2^ '• ^^- ^^- ^ Therefore DE is the fourth proportional to m, n, and p, by § 259. Q.E.D. 308. Corollary. To find the third proportional to two given lines. In the above proof take m, w, n as the given lines instead of m, n, p. 184 BOOK III. PLANE GEOMETEY Proposition XXVII. Problem: 309. To find the mean proportional between two given lines. / H \ \ 1 m ! n ! ! A m n B Given the two lines m and n. Required to find the mean proportional between m and n. Construction. Draw any line AE, and on AE take AC equal to m, and CB equal to n. On yli^ as a diameter describe a semicircle. At C erect the _L CH, meeting the circle at //. Then CH is the mean proportional between m and n. q.e.f. Proof. AC :CH=CH: CB. § 297 ( The A. from any point on a circle to a diameter is the mean proportional between the segments of the diameter.) Substituting for AC and CB their equals m and n, we have m : CH= CH : n, by Ax. 9. q.e.d. 310. Extreme and Mean Ratio. If a line is divided into two segments such that one segment is the mean proportional be- tween the whole line and the other segment, the line is said to be divided in extreme and mean ratio. E.g. the line a is divided in extreme and mean ratio, if a segment x is found such that a : X = X : a — X. The division of a line in extreme and mean ratio is often called the Golden Section. PROBLEMS OF CONSTRUCTION 185 Proposition XXVIII. Problem 311. To divide a given line in extreme and mean ratio. :>Q ^-1 ,.""nv A C B Given the line AB. Recfiired to divide AB in extreme and mean ratio. Construction. At B erect a ± BE equal to half of AB. From £^ as a center, with a radius equal to EB, describe a O. Draw AE, meeting the circle at F and G. On ^^ take ^C equal to .IF. On BA produced take A C ' equal to A G. Then AB \^ divided internally at C and externally at C in extreme and mean ratio. That is, AB : AC = AC : CB, and AB : AC = AC : C'B. q.e.f. Proof. AG:AB = A B : A F. § 302 From AG:AB = AB:AF, AG-AB:AB = AB-AF:AF. .'.AG-FG:AB = AB-AC:AC. .'. AC:AB = CB:AC. .•.AB:AC = AC:CB, by inversion, § 266. q.e.d. .'. AB:AG = AF: AB. § 266 .'. AB-\-AG: AG = AF-\-AB: AB. .\ABA-AC:AC = AF + FG'.AB. .'. CB:AC = AC: AB. .-. AB:AC = AC:C'B, by §§261, 264. q.e.d. 186 BOOK III. PLANE GEOMETRY Proposition XXIX. Pkoblem 312. Upon a given line correspondincj to a given side of a given polygon, to construct a polygon similar to the given polygon. Y I I z /i \ \ / I \ ,^ \ / --" / Given the line A^B^ and the polygon ABCDE. Required to construct on A'B\ corresponding to AB, a polygon similar to the polygon ABCDE. Construction. From A draw the diagonals AD and AC. From A ' draw A 'X, A'Y, and A 'Z, making Ax', y\ and z' equal respectively to A x, y, and «. § 232 From B' draw a line, making Z B' equal to Z B, and meeting A'X at C'. From C" draw a line, making Z D'C'B' equal to Z DCB, and meeting A'Y sX D'. From i)' draw a line, making ZE'D'C equal to ZEDC, and meeting ^'Z at E'. Then A'B'C'D'E' is the required polygon. qe.f. Proof. The A ABC and ^'^'C, the AACD and yl'C'/)', and the A ADE and ^'Z>'J5:', are similar. § 286 Therefore the two polygons are similar, by § 293. q.e.d. EXERCISES 187 EXERCISE 48 1. If a and h are two given lines, construct a line equal to ar, where x = 'Wab. Consider the special case of « = 2, ^ = 3. 2. If m and n are two given lines, construct a line equal to x, where x = V 2 mn. 3. Determine both by geometric construction and arith- metically the third proportional to the lines 1^ in. and 2 in. 4. Determine both by geometric construction and arith- metically the third proportional to the lines 4 in. and 3 in. 5. Determine both by geometric construction and arith^ metically the fourth proportional to the lines li in., 2 in.', and 21 in. 6. Determine both by geometric construction and arith- metically the mean proportional between the lines 1.2 in. and 2.7 in. 7. Eind geometrically the square root of 5. Measure the liiie and thus determine the approximate arithmetical value. 8. A map is drawn to the scale of 1 in. to 50 mi. How far apart are two places that are 2^\ in. apart on the map ? 9. Eind by geometric construction and arithmetically the third proportional to the two lines ly^^ in. and 2| in. 10. Divide a line 1 in. long in extreme and mean ratio. Measure the two segments and determine their lengths to the nearest sixteenth of an inch. 11. Divide a line 5 in. long in extreme and mean ratio. Measure the two segments and determine their lengths to the nearest sixteenth of an inch. 12. Divide a line 6 in. long in extreme and mean ratio. Measure the two segments and determine their lengths to the nearest sixteenth of an inch. The propositions on this page are taken from recent college entrance examination papers. 188 BOOK III. PLANE GEOMETRY 13. Through a given point P within a given circle to draw a chord AB ^o that the ratio AP : BP shall equal a given ratio w : n. Draw OPC so that OP-.PC = n: m. Draw CA equal to the fourth proportional to n, w, and the radius of the circle. 14. To draw two lines making an angle of 60°, and to con- struct all the circles of ^ in. radius that are tangent to both lines. 15. To draw through a given point P in the arc subtended by a chord ABsl chord which shall be bisected by AB. e^ On radius OP take CD equal to CP. Draw DE II to BA. 16. To construct two circles of radii ^ in. and 1 in. respec- tively, which shall be tangent externally, and to construct a third circle of radius 3 in., which shall be tangent to each of these two circles and inclose both of them. 17. To draw through a given external point P a secant PAB to a given circle so that the ratio PA : AB shall equal the given ratio 7)1 : n. Draw the tangent PC. Make PD . DC = m . n. PA : PC = PC : PB. 18. To draw through a given external point P a secant PAB to a given circle so that AB^ = PA X PB. 19. An equilateral triangle ABC is 2 in. on a side. To construct a circle which shall be tangent to A B at the point A and shall pass through the point C. 20. To draw through one of the points of intersection of two circles a secant so that the two chords that are formed shall be in the given ratio m : n. P^~, EXERCISES 189 21. In a circle of 3 in. radius chords are drawn through a point 1 in. from the center. What is the product of the seg- ments of these chords ? 22. The chord AB is S in. long, and it is produced through B to the point P so that PB is equal to 12 in. Find the tangent from P. 23. Two lines AB and CD intersect at 0. How would you ascertain, by measuring OA, OB, OC, and OD, whether or not the four points A, B, C, and D lie on the same circle ? 24. This figure represents an instrument for finding the centers of circular plates or sections of shafts. OC is a ruler that bisects the angle A OB, and A and OB are equal. Show that, if A and B rest on the circle, OC passes through the center, and that by drawing two lines the center can be found. 25. If three circles are tangent externally each to the other two, the tangents at their points of contact pass through the center of the circle inscribed in the triangle formed by joining the centers of the three given circles. 26. In the isosceles triangle ABC, C is a right angle, and ^C is 4 in. With A as center and a radius 2 in. a circle is described. Required to describe another circle tangent to the first and also tangent to BC at the point B. 27. Find the center of a circle of ^ in. radius, so drawn in a semicircle of radius 2 in. as to be tangent to the semi- circle itself and to its diameter 28. To inscribe in a given circle a triangle similar to a given triangle. , , - r .c 29. To draw two straight line-segments, having given their sum and their ratio. .... .. , 190 BOOK III. PLANE GEOMETRY EXERCISE 49 Review Questions 1. What is meant by ratio ? by proportion ? 2. If a'.b=^c: d, write four other proportions involving these quantities. 3. If a : b = c : d, is it true in general that a-{-l:b -\-l z=c:d? Is it ever true ? 4. When is a line divided harmonically ? The bisectors of what angles of a triangle divide the opposite side harmonically ? 5. What are the two conditions necessary for the similarity of two polygons ? 6. Are two mutually equiangular triangles similar ? Are two mutually equiangular polygons always similar ? 7. Are two triangles similar if their corresponding sides are proportional ? Are two polygons always similar if their corresponding sides are proportional ? 8. If two triangles have their sides respectively parallel, are they similar ? Is this true of polygons in general ? 9. If two triangles have their sides respectively perpendicu- lar, are they similar ? Is this true of polygons m general ? 10. Complete in two ways : The perimeters of two similar polygons have the same ratio as any two corresponding ••• . 11. If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse, state three geometric truths that follow. 12. If two secants intersect outside, on, or within a circle, what geometric truth follows ? "13. How would you proceed to divide a straight line into seven equal parts ? - 14. How. would you proceed to find the square root of 7 by measuring the length of a line ? ' ;. ■ BOOK IV AREAS OF POLYGONS 313. Unit of Surface. A square the side of which is a unit of length is called a unit of surface. Thus a square that is 1 inch long is 1 square inch, and a square that is 1 mile long is 1 square mile. If we are measuring the dimensions of a room in feet, we measure the surface of the floor in square feet. In the same way we may measure the page of this book in square inches and the area of a state in square miles. 314. Area of a Surface. The measure of a surface, expressed in units of surface, is called its area. If a room is 20 feet long and 15 feet wide, the floor contains 300 square feet. Therefore the area of the floor is 300 square feet. Usually the two sides of a rectangle are not commensurable, although by means of frac- tions we may measure them to any required degree of approximation. The incommensurable cases in theorems like Prop. I of this Book may be omitted without interfering with the sequence of the course. 315. Equivalent Figures. Plane figures that have equal areas are said to be equivalent. In propositions relating to areas the words rectangle^ triangle^ etc., are often used for area of rectangle^ area of triangle^ etc. Since congruent figures may be made to coincide, congruent figures are manifestly equivalent. Because their areas are equal, equivalent figures are frequently spoken of as equal figures. The symbol = is used both for " equivalent " and for " congruent," the sense determining which meaning is to be assigned to it. Occasionally these symbols are used : = , = , or = for congruent, = for equal, and =o for equivalent. Since the word congruent means " identically equal," the word equal is often used to mean ''equivalent." 191 192 BOOK IV. PLANE GEOMETRY Proposition I. Theorem 316. Two rectangles having equal altitudes are to each other as their bases. D x B A X Given the rectangles -AC and AF^ having equal altitudes AD. To prove that \I}AC:\I\AF = base AB : base AE. Case 1. When AB and AE are commensurable. Proof. Suppose AB and AE have a common measure, as AX. Suppose AX '\?> contained m times in AB and n times in AE. Then AB:AE = m:n. (For m and n are the numerical measures of AB and AE.) Apply ^X as a unit of measure to ^^ and AE, and at the several points of division erect Js. These Js are all ± to the upper bases, § 97 and these Js are all equal. § 128 Since to each base equal to AX there is one rectangle, .'. □ .4C is divided into m rectangles, and □ .4F is divided into n rectangles. § 119 These rectangles are all congruent. § 133 .-, ]Z\AC'.L^AF = m:n. .'. {3 AC '.n^AF^^AB'.AE, by Ax. 8. q.e.d. In this proposition we again meet the incommensurable case, as on pages 116 and 157. This case is considered on page 193 and may be omitted without destroying the sequence of the propositions. AREAS OF POLYGONS 193 Case 2. When AB and AE are incommensurable. D D Proof. Divide A E into any number of equal parts, and apply one of these parts to AB as many times as AB will contain it. Since AB and AE are incommensurable, a certain number of these parts will extend from A to some point P, leaving a re- mainder PB less than one of them. Draw PQ _L to AB. □ .4Q AP \Z\AF~ AE' Then Case 1 By increasing the number of equal parts into which AE is divided we can diminish the length of each, and therefore can make PB less than any assigned positive value, however small. Hence PB approaches zero as a limit, as the number of parts of AE is indefinitely increased, and at the same time the cor- responding □ PC approaches zero as a limit. § 204 Therefore AP approaches yl^ as a limit, and D^Q ap- proaches \Z\ AC as a limit. .*. the variable — — approaches — — as a limit, ACt A. E and the variable 7=— — approaches 7=— — as a limit. A P CH A Q But — — is always equal to ^^\ „ ? as AP varies in value and \3AF approaches AB as a limit. n\AC AB ^ , ^_ Case 1 Q.E.D. 317. Corollary. Two rectangles having equal bases are to each other as their altitudes. 194 BOOK IV. PLANE GEOMETRY Proposition II. Theorem 318. Tivo rectangles are to each other as the j^roducts of their bases hy their altitudes. R b U b Given the rectangles i? and R\ having for the numerical measure of their bases b and h\ and of their altitudes a and a! respectively. To prove that -— = -— • R a'b' Proof. Construct the rectangle S, with its base equal to that of R, and its altitude equal to that of R\ R_a ' R'~b'' Then and §317 §316 Since we are considering areas, we may treat R, S, and S' as numbers and take the products of the corresponding members of these equations. ,-, ^ § 272 We therefore have n_ R' ab by Ax. 3. Q.E.D. 319. Products of Lines. When we speak of the product of a and b we mean the product of their numerical values. It is possible, however, to think of a line as the product of two lines, by changing the definition of multiplication. Thus in this fig\ire in which two parallels are cut by two intersecting trans- versals, we have l:a = b : x. Therefore x — ah. In the same way we may find xc^ or abc^ the product of three lines. AREAS OF POLYGONS Proposition III. Theorem 195 320. The area of a rectangle is equal to the product of its base hy its altitude. Q Given the rectangle /?, having for the numerical measure of its base and altitude b and a respectively. To prove that the area of R — ah. Proof. Let U be the unit of surface. § 313 Then ab 1x1 ah. §318 R But J- = the number of units of surface in R, i.e. the area oiR. §314 .". the area of R = ab, by Ax. 8. q.e.d. 321. Practical Measures. When the base and altitude both contain the linear unit an integral number of times, this propo- sition is rendered evident by dividing the rectangle into squares, each equal to the unit of surface. Thus, if the base contains seven hnear units and the altitude four, the rectangle may be divided into twenty-eight squares, each equal to the unit of surface. Practi- cally this is the way in which we conceive the measure of all rectangles. Even if the sides are incommensurable, we cannot determine this by any measuring instrument. If they seem to be incommensurable with a unit of a thousandth of an inch, they might not seem to be incommensurable with a unit of a millionth of an inch. 196 BOOK IV. PLANE GEOMETRY EXERCISE 50 1. A square and a rectangle have equal perimeters, 144 yd., and the length of the rectangle is five times the breadth. Compare the areas of the square and rectangle. 2. On a certain map the linear scale is 1 in. to 10 mi. How many acres are represented by a square | in. on a side ? 3. Find the ratio of a lot 90 ft. long by 60 ft. wide to a field 40 rd. long by 20 rd. wide. 4. Find the area of a gravel walk 3 ft. 6 in. wide, which surrounds a rectangular plot of grass 40 ft. long and 25 ft. wide. Make a drawing to scale before beginning to compute. 5. Find the number of square inches in this cross section of an L beam, the thickness being J in. ' 6. What is the perimeter of a square field that contains exactly an acre ? '>^-2%-^ 7. A machine for planing iron plates planes a space ^ in. wide and 18 ft. long in 1 min. How long will it take to plane a plate 22 ft. 6 in. long and 4 ft. 6 in. wide, allowing 51 min. for adjusting the machine ? 8. How many tiles, each 8 in. square, will it take to cover a floor 24 ft. 8 in. long by 16 ft. wide ? 9. A rectangle having an area of 48 sq. in. is three times as long as wide. What are the dimensions ? 10. The length of a rectangle is four times the width. If the perimeter is 60 ft., what is the area? 11. From two adjacent sides of a rectangular field 60 rd. long and 40 rd. wide a road is cut 4 rd. wide. How many acres are cut off for the road ? 12. From one end of a rectangular sheet of iron 10 in. long a square piece is cut off leaving 25 sq. in. in the rest of the sheet. How wide is the sheet ? AREAS OF POLYGONS Proposition IV. Theorem 197 322. Tlie area of a parallelogram is equal to the product of its base by its altitude. YD X G A b B A b B Given the parallelogram ABCD^ with base h and altitude a. To prove that the area of the CJABCD = ah. Proof. From B draw BX 1. to CD or to CD produced, and from A draw ^ F J_ to CD produced. Then ABXY is a rectangle, with base b and altitude a. Since AY= BX, 2indAD = BC, § 125 .-.the rt. A ADY and BCX are congruent. § 89 From ABC Y take the A BCX; the \Z1ABXY is left. From ABC Y take the A ADY; the EJABCD is left. .-. n\ABXY= CJABCD. Ax. 2 But the area of the □ A BX Y = ah. § 320 .*. the area of the CJABCD = ah, by Ax. 8. q.e.d. 323. C'oKOLLARY 1. P aralUlograms having equal bases and equal altitudes are equivalent. 324. Corollary 2. Parallelograms having equal bases are to each other as their altitudes; parallelograms having equal altitudes are to each other as their bases ; any two jmrallelo- grams are to each other as the products of their bases by their altitudes. This was regarded as veiy interesting by the ancients, since an ignorant person might think it impossible that the areas of two parallelograms could remain the same although their perimeters differed without limit. 198 BOOK IV. PLANE GEOMETRY Proposition V. Theorem 325. The area of a triangle is equal to half the product of its base by its altitude. A b B Given the triangle ABC, with altitude a and base b. To prove that the area of the A ABC = ^ab. Proof. With AB and BC SiS adjacent sides construct the parallelogram A BCD. § 238 Then A ABC = i EJABCD. § 126 But the area of the EJABCD = ah. § 322 .*. the area of the A ABC = \ ab, by Ax. 4. q.e.d. 326. Corollary 1. Triangles having equal bases and equal altitudes are equivalent. 327. Corollary 2. Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are to each other as their bases ; any two triangles are to each other as the products of their bases by their altitudes. Has this been proved for rectangles ? What is the relation of a triangle to a rectangle of equal base and equal altitude ? What must then be the relations of triangles to one another ? 328. Corollary 3. The product of the sides of a right tri- angle is equal to the product of the hypotenuse by the altitude from the vertex of the right angle. How is the area of a right triangle found in terms of the sides of the right angle ? in terms of the hypotenuse and altitude ? How do these results compare ? AREAS OF POLYGONS 199 Proposition VI. Theorem 329. The area of a trapezoid is equal to half the product of the sum of its bases by its altitude. A b B Given the trapezoid ABCD^ with bases b and V and altitude a. To prove that the area of ABCD ='^a(h-\- h'}. Proof. Di-aw the diagonal A C. Then the area of the AABC = ^ ah, and the area of the A A CZ) = ^ ah'. § 325 .*. the area of ABCD = ^ a (^ + h'), by Ax. 1. q.e.d. 330. Corollary. The area of a trapezoid is equal to the product of the line joining the mid-points of its nonparallel sides hy its altitude. How is the line joining the mid-points of the nonparallel sides related to the sum of the bases (§ 137) ? 331. Area of an Irregular Polygon. The area of an irregular polygon may be found by dividing the polygon into triangles, and then finding the area of each of these triangles separately. A common method used in land sur- veying is as follows : Draw the longest diagonal, and let fall perpendiculars upon this diagonal from the other ver- tices of the polygon. The sum of the right triangles, rectahglep, and trapezoids is equivalent to the polygon. 200 BOOK IV. PLANE GEOMETEY EXERCISE 51 Find the areas of the parallelograms whose bases and altitudes are respectively as follows : 1. 2.25 in., 1 J in. 3. 2.7 ft., 1.2 ft. 5. 2 ft. 3 in., 7 in. 2. 3.44 in., li in. 4. 5.6 ft., 2.3 ft. 6. 3 ft. 6 in., 2 ft. Find the areas of the triangles whose bases and altitudes ai'e respectively as follows : 7. 1.4 in., 11 in. 9. 6^ ft., 3 ft. 11. 1 ft. 6 in., 8 in. 8. 2.5 in., 0.8 in. 10. 5.4 ft., 1.2 ft. 12. 3 ft. 8 in., 3 ft. Find the areas of the trapezoids ivhose bases are the first two of the following numbers, and whose altitudes are the third numbers : 13. 2 ft., 1 ft., 6 in. 15. 3 ft. 7 in., 2 ft., 14 in. 14. 2^ ft., 11 ft., 9 in. 16. 5 ft. 6 in., 3 ft., 2 ft. Find the altitudes of the parallelograms whose areas and bases are respectively as follows : 17. 10 sq. in., 5 in. 19. 28 sq. ft., 7 ft. 21. 30 sq. ft., 12 ft. 18. 6sq. in., 6in. 20. 27 sq. ft., 6 ft. 22. 80 sq. in., 16 in. Find the altitudes of the triangles whose areas and bases are respectively as follows : 23. 49 sq. in., 14 in. 25. 50 sq. ft., 10 ft. 27. 110 sq. yd., 10 yd. 24. 48 sq. in., 12 in. 26. 160 sq. ft., 20 ft. 28. 176 sq. yd., 32 yd. Find the altitudes of the trapezoids whose areas and bases are respectively as follows : 29. 33 sq. in., 5 in., 6 in. 31. 13 sq. ft., 9 ft., 5 ft. 30. 15 sq. in., 4 in., 6 in. 32. 70 sq. yd., 9 yd., 11 yd. ^. , AREAS OF POLYGONS 201 Proposition YII. Theorem 332. The areas of two triangles that have an angle of the one equal to an angle of the other are to each other as the 2^roducts of the sides including the equal angles. Given the triangles ABC and ADE^ with the common angle A, ^ , , A ABC ABxAC To prove that -7— rrrr. = ~7T^ 7~r,- ^ AADE ADxAE Proof. Draw BE. Then A ABC AC A ABE AE A ABE AB ^ ^^„ {Triangleff having equal altitudes are to each other as their bases.) Since we are considering numerical measures of area and length, we may treat all of the terms of these proportions as, numbers. Taking the product of the first members and the product of the second members of these equations, we have AABE X A ABC AB x AC A ADE X A ABE AD X AE That is, by canceling AABE, we have the proportion Ax. 3 AABC ABxAC = O.E.D AADE ADXAE 202 BOOK IV. PLANE GEOMETRY Proposition VIII. Theorem 333. The areas of tiuo similar triangles are to each other as the squares on any two corresjjonding sides. A B A' Given the similar triangles ABC and A'B'C\ ^ ^, ^ A ABC AB' To prove that .fp,^> = ■ 2 • Proof. Since the triangles are similar, Given .\AA=AA'. §282 _, A ABC ABxAC ^^^^ aJ^B^'^A'B'xA'C' ^^^^ ( The areas of two triangles that have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.) . AABC _ AB_ AC_ Ihatis, AA'B'C'- A'B'^A'C' But -^ = ~ • §282 A'B' A'C {Similar polygons have their corresponding sides proportional.) Substituting — — , for its equal 7777,' we have A J:> A Ly AABC _ AB_ AB_ AA'B'C'- A'B' ^ A'B'' ^^ AABC AB^ AREAS OF POLYGONS 203 Proposition IX. Theorem 334. The areas of two similar polygons are to each other as the squares on any tioo corresponding sides. Given the similar polygons ABODE and A'B^C'D'E\ of area 5 and s' respectively. To prove that s:s' = AB'':A'B'\ Proof. By drawing all the diagonals from any correspond- ing vertices .4 and A', the two similar polygons are divided into similar triangles. AADE ad' AACB AC'' A ABC AA'D'E' Ji^r^ AA'C'D' jJc' AADE AACD That is, AA'D'E' AA'C'D' AADE-{-AACD-\-AABC •' AA'B'C _ A ABC ~ AA'B'C' A ABC AA'D'E' + A A'C'D' + A A'B'C AA'B'C :2 -TT^2 §292 ab' A'B'' • §333 Ax. 8 ab' §269 s:s' = AB :A'B' , by Ax. 11. Q.E.D. 335. Corollary 1. The areas of tivo similar polygons are to each other as the squares on any two corresponding lines. 336. Corollary 2. Corresponding sides of hvo similar poly- gons have the same ratio as the square roots of the areas. 204 BOOK IV. PLANE GEOMETKY Proposition X. Theorem 337. Tlie square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other tivo sides. / R X S Given the right triangle ABC^ with AS the square on the hypote- nuse, and BN^ CQ the squares on the other two sides. To prove that AS=BN-\-CQ. Proof. Draw CX through C II to ^.S". § 233 Draw CR and BQ. Since A c and x are rt. A, the Z PCB is a straight angle, § 34 and the line PCB is a straight line. Similarly, the line A CN is a straight line. In the A ARC and ABQ, AR = AB, AC = AQ, and ARAC = ZBAQ. {Each is the sum of a rt. Z and the Z BAC.) .'. A ARC is congruent to A ABQ. Furthermore the □ AX is double the A ARC. {They have the same base AR, and the same altitude RX.) §43 ^65 Ax. 1 §68 §325 KUMERICAL PROPERTIES OF FIGURES 205 Again the square CQ is double the AABQ. § 325 {They have the same base AQ, and the same altitude AC.) .'. the \I2AX is equivalent to the square CQ. Ax. 3 In like manner, by drawing CS and AM, it may be proved that the rectangle BX is equivalent to the square BN. Since square AS = nBX-^[D AX, Ax. 11 .-. AS = BN-{- CQ, by Ax. 9. q.e.d. The first proof of this theorem is usually attributed to Pythagoras (about 525 B.C.), although the truth of the proposition was known earlier. It is one of the most important propositions of geometry. Various proofs may be given, but the on6 here used is the most common. This proof is attrib- uted to Euclid (about 300 b.c), a famous Greek geometer. 338. Corollary 1. The square on either side of a right triangle is equivalent to the difference of the square on the hypotemise and the square on the other side. 339. Corollary 2. The diagonal and a side of a square are incommensurable. jy ^ For AC^ = AB^ + BC^ = 2 AB^. .: AC = ABV2. Since V2 may be carried to as many decimal places as we please, but cannot be exactly expressed as a rational fraction, it has no common measure with 1. That is, AC :AB = V2, an incommensurable number. 340. Projection. If from the extremities of a line-segment perpendiculars are let fall upon another line, the segment thus cut off is called the projection of the first line upon the second. Thus C^iy is the projection of CD upon AB, or V is the projection of I upon AB. In general it is convenient to designate by the small letter a the side of a triangle opposite ZA, and so for the other sides; to designate the projection of a by a'; and to designate the height (altitude) by A. A c^ d' 206 BOOK IV. PLANE GEOMETRY EXERCISE 52 Given the sides of a right triangle as follotvs^ find the hypotenuse to two decimal places: 1. 30 ft., 40 ft. 3. 20 ft., 30 ft. 5. 2 ft. 6 in., 3 ft. 2. 45 ft., 60 ft. 4. 1.5 in., 2.5 in. 6. 3 ft. 8 in., 2 ft. Given the hypotenuse and one side of a right triangle as follows, find the other side to two decimal places : 7. 50 ft, 40 ft. 9. 10 ft., 6 ft. 11. 3 ft. 4 in., 2 ft. 8. 35 ft., 21 ft. 10. 1.2 in., 0.8 in. 12. 6 ft. 2 in., 5 ft. 13. A ladder 38 ft. 6 in. long is placed against a wall, with its foot 23.1 ft. from the base of the wall. How high does it reach on the wall ? 14. Find the altitude of an equilateral triangle with side s. 15. Find the side of an equilateral triangle with altitude h. 16. The area of an equilateral triangle with side s is i s^ V3. 17. Find the length of the longest chord and of the shortest chord that can be drawn through a point 1 ft. from the center of a circle whose radius is 20 in. 18. The radius of a circle is 5 in. Through a point 3 in. from the center a diameter is drawn, and also a chord perpen- dicular to the diameter. Find the length of this chord, and the distance (to two decimal places) from one end of the chord to the ends of the diameter. 19. In this figure the angle C is a >/l\ right angle. From the relations AC = y^ \ \ AB X AF (§ 294) ^d C^= AB x BF, ^^ X I \ ^ show that lc'+ c]b% Zb^ ^ 20. If the diagonals of a quadrilateral intersect at right angles, the sum of the squares on one pair of opposite sides is equivalent to the sum of the squares on the other pair. NUMERICAL PROPERTIES OF FIGURES 207 Proposition XI. Theorem 341. In any triangle the square on the side opposite an acute angle is equivalent to the sum of the squares on the other two sides diminished hy ttvice the product of one of those sides hy the projection of the other upon that side. c Fig. 1 Given the triangle ABC^ A being an acute angle, and a' and V being the projections of a and b resi)ectively upon c. To prove that d^ = ¥-\-c^—2 h'c. Proof. If D, the foot of the _L from C, falls upon c (Fig. 1), a' = c-b'. If D falls upon c produced (Fig. 2), a' = b'-c. In either case, by squaring, we have a'2 = Z»'2_f_^2_2^'c. Ax. 5 Adding A^ to each side of this equation, we have h^-^a'^ = h^-{-b'^ + c^-2b'c. Ax. 1 But h^ + «'2 = a^, and h^ + h"" = b\ § 337 Putting a^ and b'^ for their equals in the above equation, we have a^ = b^^c^-2b'c, by Ax. 9. Q.E.D. 208 BOOK IV. PLANE GEOMETRY Proposition XII. Theorem 342. In any obtuse triangle the square on the side opposite the obtuse angle is equivalent to the sum of the squares on the other tivo sides increased by tivice the product of one of those sides by the projection of the other upon that side. Given the obtuse triangle ABC^ A being the obtuse angle, and a' and b' the projections of a and h respectively upon c. To prove that a^ = b^ + e'-\-2 b'c. Proof. a' = b' -\-c. Squaring, a''' = b'"" + e" -{- 2 b'c. Adding h^ to each side of this equation, we have 7,2 -f ,,'2 =h^^ z,'2 4_ ^.2 _^ 2 b'c. But h^ + a'^ = a^, and Ii^ + b'^ = b\ Ax. 11 Ax. 5 Ax. 1 §337 Putting (i^^iYidi V^ for their equals in the above equation, we have a^ = b^^^-^2b'c,\^j Ky.. 9. Q.E.D. Discussion. By the Principle of Continuity the last three theorems may be included in one theorem by letting the A A change from an acute angle to a right angle and then to an obtuse angle. Let the student explain. The last three theorems enable us to compute the altitudes of a tri- angle if the three sides are known ; for in Prop. XII we can find 6', and from h and V we can find h. NUMERICAL PROPERTIES OF FIGURES 209 EXERCISE 53 Find the lengths^ to two decimal places, of the diagonals of the squares whose sides are : 1. 7 in. 2. 10 in. 3. 9.2 in. 4. 1 ft. 6 in. 5. 2 ft. 3 in. Find the lengths, to two decimal places, of the sides of the squares whose diagonals are : 6. 4 in. 7. 8 in. 8. 5 ft. 9. VE in. 10. 2 ft. 6 in. 11. The minute hand and hour hand of a clock are 6 in. and 4^ in. long respectively. How far apart are the ends of the hands at 9 o'clock ? 12. A rectangle whose base is 9 and diagonal 15 has the same area as a square whose side is x. Find the value of x. 13. A ring is screwed into a ceiling in a room 10 ft. high. Two rings are screwed into the floor at points 5 ft. and 12 ft. from a point directly beneath the one in the ceiling. Wires are stretched from the ceiling ring to each floor ring. How long are the wires ? (Answer to two decimal places.) 14. The sum of the squares on the segments of two perpen- dicular chords is equivalent to the square on the diameter of the circle. If AB, CD are the chords, draw the diameter BE, and draw AC, ED, BD. Prove that AC = ED. 15. The difference of the squares on two sides of a triangle is equivalent to the difference of the squares on the segments of the third side, made by the perpendicular on the third side from the opposite vertex. 16. In an isosceles triangle the square on one of the equal sides is equivalent to the square on any line drawn from the vertex to the base, increased by the product of the segments of the base. 210 BOOK IV. PLANE GEOMETRY Proposition XIII. Theorem 343. The sum of the squares on tioo sides of a tri- angle is equivalent to tivice the square o?i half the third side, increased hy tivice the square on the median upon that side. The difference of the squares on tioo sides of a tri- angle is equivalent to tivice the j^^oduct of the third side hy the projection of the median upon that side. Given the triangle ABC^ the median m, and mJ the projection of m upon the side a. Also let c be greater than b. To prove that 1. c' + 6' = 2 .RE' + 2 w%- 2. c'-l>' = 2am'. ■ Proof. The Z AMB is obtuse, and the Z CMA is acute. § 116 Since c>b, M lies between B and D. § 84 Then and c^ = BM ' + m^ -\-2BM-m', IP' = MC^ -\-m^-2MC ' m'. §342 §341 Adding these equals, and observing that BiM = MC, we have c'i_^p = 2 BM^ + 2 m^. Ax. 1 Subtracting the second from the first, we have c^ — b^ = 2 am', by Ax. 2. q.e.d. Discussion. Consider the proposition when c = h. This theorem may be omitted without interfering with the regular sequence. It enables us to compute the medians when the three sides are known. EXERCISES 211 EXERCISE 54 1. To compute the area of a triangle in terms of its sides. c A C B D At least one of the angles ^ or B is acute. Suppose A is acute. In the A A DC, h^ = b"-AD^. Why ? In the A ABC, a^ = ¥ -{- c^ - 2c x AD. Why ? 62 4- c2 - a2 Therefore AD Hence h^ = b- 2c (62 4. c2 _ a2)2 4 ^2^2 _ (52 4. c2 _ a2)2 4c2 4c2 _ (2 6c + 62 ^ (.2 - a2) (2 6c - 62 _ c2 + a2) " ~ 4c2 _ {(6 + c)2 - a2} {a2 - (6 - c)2} ~ 4c2 _ (g + 6 + c) (6 + c — g) (g + 6 — c) (g — 6 + c) - 4c2"' Let g+6 + c = 2 5, where 5 stands for semiperimeter. Then 6 + c — g = g + 6 + c — 2g = 2s — 2g = 2(s — g). Similarly • g + 6 — c = 2 (s — c), and g-6 + c = 2(s-6). ^„ 2 s X 2 (.s - g) X 2 (8 - 6) X 2 (s - c) Hence h^ = ^^ ^^ ^^ • 4c2 By simplifying, and extracting the square root, 2 / hz= -Vs{s — a) (s — 6) (s — c). Hence the area = lch= Vs [s — a) {s — 6) {s — c). For example, if the sides arc 3, 4, and 6, area = V6(6 - 3) (6 - 4) (6 - 5) = V6 • 3 • 2 = 6. 212 BOOK IV. PLANE GEOMETRY If Ex. 1 has been studied, find the areas, to two decimal places, of the triangles whose sides are : 2. 4, 5, 6. 4. 6, 8, 10. 6. 7, 8, 11. 8. 1.2, 3, 2.1. 3. 5, 6, 7. 5. 6, 8, 9. 7. 9, 10, 11. 9. 11, 12, 13. 10. To compute the radius of the circle circumscribed about a triangle in terms of the sides of the triangle. (Solve only if § 305 and Ex. 1 have been taken.) Let CD be a diameter. By § 305, what do we know about the products CA x BC and CD x CP ? What does this tell us of ab and 2 r • CP, r be- ing the radius ? X^ ^^^^^^ From Ex. 1, what does CP equal in terms of / ^^-^^^^/ i ^ the sides ? -^[ ^ c — 7^ — h~^^ Is it therefore possible to show that I / ^-^^j abc „ \ /'"^^ J 4 Vs (s — a) (8 — 6) (s — c) .dV_ ^^ If Exs. 1 and 10 have been studied, compute the radii, to two decimal places, of the circles circumscribed about the triangles whose sides are : 11. 3, 4, 5. 12. 27, 36, 45. 13. 7, 9, 11. 14. 10, 11, 12. 15. To compute the medians of a triangle in terms of its sides. Omit if § 343 has not been taken. What do we know about a^ + 6^ as compared with 2 m^ -j- 2 ( - J ? From this relation show that If Ex. 15 has been studied, compute the three medians, to two decimal places, of the triangles whose sides are : 16. 3, 4, 5. 17. 6, 8, 10. 18. 6, 7, 8. 19. 7, 9, 11. 20. If the sides of a triangle are 7, 9, and 11, is the angle opposite the side 11 right, acute, or obtuse ? EXERCISES 213 1^ I -I . H .H \C 21. The square constructed upon the sum of two lines is equivalent to the sum of the squares constructed upon these two lines, increased by twice the rectangle of these lines. Given the two lines AB and BC, and AC their sum. Construct the squares AKGC and ADEB upon AC and AB respec- tively. Produce BE and DE to meet KG and CG in H ^ B c and F respectively. Then we have the square EHGF, with sides each equal to BC. Hence the square AKGC is the sum of the squares ADEB and EHGF, and the rectangles DKHE and BEFC. This proves geometrically the algebraic formula (a + 6)2 = a2 + 2 a6 + b^. 22. The square constructed upon the difference of two lines is equivalent to the sum of the squares constructed upon these two lines, diminished by twice their rectangle. Given the two lines AB and AC, and BC their dif- ference. Construct the square AGFB upon AB, the square A CKH upon A C, and the square CDEB upon BC. Produce ED to meet AG in L. The dimensions of the rectangles LGFE and HLDK are AB and AC, and the square CDEB is the difference between the whole figure and the sum of these rectangles. This proves geometrically the algebraic formula (a _ 6)2 = a2 - 2 ab + b^. 23. The difference between the squares constructed upon' two lines is equivalent to the rectangle of the sum and differ- ence of these lines. / k Given the squares ^BD J? and CBFG, constructed upon AB and BC. The difference between these squares is the polygon ACGFDE, which is composed of the rectangles ACHE and GFDH. Produce AE and CH to / and K respectively, making EI and HK each equal to BC, and draw IK. The difference be- ^ tween the squares ABDE and CBFG is then equiva- lent to the rectangle ACKI, with dimensions AB + BC, and AB This proves geometrically the algebraic formula a2- 62 = (a + 6) (a -6). . 1> \ H G BC. 214 BOOK IV. PLANE GEOMETRY Proposition XIV. Problem 344. To construct a square equivalent to the sum of tivo given squai^es. Given the two squares, i? and i?'. Required to construct a square equivalent to E -\- E'. Construction. Construct the rt. Z.4. § 228 On the sides of Z .4, take AB, or c, equal to a side of R', and AC, or b, equal to a side of R, and draw BC, or a. Construct the square S, having a side equal to BC. Then ^ is the square required. q.e.f. Proof. a-=Z;'--|-6'2. §337 {The square on the hypotenuse of a rt. A is equivalent to the sum of the squares on the other two sides.) .'. S = R-{-R', by Ax. 9. q.e.d. 345. Corollary 1. To construct a square equivalent to the difference of two given squares. We may easily reverse the above construction by first dravi^ing .c. then erecting a ± at ^, and then with a radius a fixing the point C. 346. Corollary 2. To construct a square equivalent to the Slim of three given squares. If a side of the third square is d, we may erect a perpendicular from C to the line J5C, take CD equal to d, and join D and B. Discussion. It is evident that we can continue this process indefi- nitely, and thus construct a square equivalent to the sum of any number of given squares. PKOBLEMS OF CONSTKUCTION 215 Proposition XV. Problem 347. To construct ci j^ohjgon similar to two given similar polygons and equivalent to their sicrn. R" Given the two similar polygons R and i?'. Required to construct a polygon similar to R and R\ and equivalent to R -\- R\ Construction. Construct the rt AO. § 228 Let s and s' be corresponding sides of R and R\ On the sides of /.O, take OX equal to s', and OY equal to s. Draw XY, and take s" equal to XY. Upon s", corresponding to s, construct W similar to 11. § 312 Then it" is the polygon required. q.e.f. Proof. OF^ -\-'0X'^ = XY^. § 337 Putting for OY, OX, and XF their equals s, s', and s", we have But and By addition, R"~s"^'' R"~s"^' R-^R' s'-\- R' 1. :. R" = R + PJ, by Ax. 3. Ax. 9 §334 Ax. 1 Q.E.D. 216 BOOK IV. PLANE GEOMETRY Pkoposition XVI. Problem 348. To construct a triangle equivalent to a given polygon. Given the polygon ABCDEF. Required to construct a triangle equivalent to ABCDEF. Construction. Let B, C, and D be any three consecutive vertices of the polygon. Draw the diagonal DB. From V draw a line II to DB. § 233 Produce AB to meet this line at Q, and draw DQ. Again, draw EQ, and from D draw a line II to EQ, meeting AB produced at R, and draw ER. In like manner continue to reduce the number of sides of the polygon until we obtain the A EPR. Then A £JP/? is the triangle required. q.e.f. Proof. The polygon AQDEF has one side less than the polygon ABCDEF. Furthermore, in the two polygons, the ipSii'tABDEF is common, and the A BQD = A BCD. § 326 (For the base DB is common, and their vertices C and Q are in the line CQ II to the base.) .-. AQDEF = ABCDEF. Ax. 1 In like manner it may be proved that AREF= AQDEF, and EPR = AREF. Q.e.d. PROBLEMS OF CONSTRUCTION 217 Proposition XVII. Problem 349. To construct a square equivalent to a given parallelogram. p N M A h B Given the parallelogram ABCD. Required to construct a square equivalent to the EJABCD. Construction. Upon any convenient line take NO equal to a, and OM equal to b, the altitude and base respectively of O ABCD. Upon NM as a diameter describe a semicircle. At O erect OP _L to NM, meeting the circle at P. § 228 Construct the square S, having a side equal to OP. Then S is the square required. q.e.f. Proof. That is, But and NO :OP = OP: OM. .'. 0P^=N0X0M. OF''=ah. S = 0P\ §297 §261 Ax. 9 EJABCD = ab. §322 .-. S=CJABCD, by Ax. 9. q.e.d. 350. Corollary 1. To construct a square equivalent to a given triangle. Take for a side of the square the mean proportional between the base and half the altitude of the triangle. 351. Corollary 2. To construct a square equivalent to a given polygon. First reduce the polygon to an equivalent triangle, and then construct a square equivalent to the triangle. 218 BOOK IV. PLANE GEOMETRY Proposition XVIII. Problem 352. To construct a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line. A Q B Given the square S, and the line AB. Required to construct a O equivalent to S^ with the sum of its base and altitude equal to AB. Construction. Upon AB as a diameter describe a semicircle. At A erect AC A- to AB and equal to a side of the given square S. § 228 Draw CD II to AB, cutting the circle at P. § 233 Draw PQ _L to AB. § 227 Then any CJ, as P, having AQ for its altitude and QB for its base is equiva lent to .S. Q.E.F. Proof. AQ:PQ = PQ: QB. §297 .'.Pq''=AQx QB. §261 Furthermore PQ is II toC^. §95 .\PQ = CA. §127 .'.PQ^ = CA\ Ax. 5 .•..4QX QB=CA\ Ax. 8 But P = AQx QB, §322 and S=zCA\ §320 .'. P=S, by Ax. 8. Q.E.D. Thus is solved geometrically the algebraic problem, given x + y = a, xy = 6, to find x and y. PKOBLEMS OF CONSTRUCTION 219 Pkoposition XIX. Problem 353. To construct a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line, c .E^ -\B Given the square S, and the line AB. Required to construct a O equivalent to S, ivith the differ- ence of its base and altitude equal to AB. Construction. Upon .45 as a diameter describe a circle. From A draw A C\ tangent to the circle, and equal to a side of the given square S. Through the center of the circle draw CD intersecting the circle at E and D. Then any O, as P, having CD for its base and CE for its altitude, is equivalent to aS. q. e. f. Proof. CD : CA = CA : CE. § 302 .'.CA^ = CDxCE, §261 and the difference between CD and CE is the diameter of the circle, that is, AB. But P=CDx CE, §322 and S=CA''. §320 .'. P= S,hy Ax. 8. Q. E. D. Thus is solved geometrically the algebraic problem, given x — y = a, xy = &, to find x and y. 220 BOOK IV. PLANE GEOMETRY Proposition XX. Problem 354. To construct a j)olygon similar to a given poly- gon and equivalent to another given polygon. Given the polygons P and Q. Required to construct a polygon similar to P and equiva- lent to Q. Construction. Construct squares equivalent to P and Q, § 351 and let m and n respectively denote their sides. Let s be any side of P. Find s\ the fourth proportional to m^ n, and s. § 307 Upon s\ corresponding to s, construct a polygon P' similar to the polygon P. § 312 Then P' is the polygon required. q.e.f. Proof. Since m : n = s : s', Const. .•.m^:7i' = s^:s'\ §270 But P = 711^, and Q = n^. Const. .'.P:Q = s'':s'\ Ax. 9 But P'.P^ = s' : s'l § 334 {The areas of two similar polygons are to each other as the squares on any two corresponding sides.) ,'. P:Q = P:P'. Ax. 8 .'.P'=Q. §263 .*. P', being similar to P, is the polygon required, q.e.d. PROBLEMS OF CONSTRUCTION 221 Proposition XXI. Problem 355. To construct a square which shall have a given ratio to a given square. s D #-^. / Given the square S, and the ratio — . m Required to construct a square which shall he to S as n is to m. Construction. Take AB equal to a side of S, and draw /IF, making any convenient angle with AB. On A Y take AE equal to m units and EF equal to n units. Draw EB. From F draw a line II to EB, meeting AB produced at C. § 233 On .4 C as a diameter describe a semicircle. At B erect BD J_ to ^ C, meeting the semicircle at D. Then BD is a side of the square required Proof. Denote AB hj a, BC hj b, and BD by x. §228 Q.E.F. Then But By inversion, a : x^x : b. . a:b = a^ :x\ a\h — m : n. a^ :x^ = m : n. x^ :a^ = n: Tfl. §297 §271 §273 Ax. 8 §266 Hence the square on BD will have the same ratio to S as n has to m. q.e.d. 222 BOOK IV. PLANE GEOMETRY Proposition XXII. Problem 356. To construct a jjolycjon similar to a given j)oly- gon and having a given ratio to it. Given the polygon P and the ratio — • 772 Required to construct a polygon similar to P, which shall he to P as n is to m. Construction. Let s be any side of P. Draw a line ,s-', such that the square on *•' shall be to the square on ^ as t^ is to m. § 355 Upon *•' as a side corresponding to s construct the polygon P' similar to P. § 312 {Upon a given line corresponding to a given side of a given polygon^ to construct a polygon similar to the given polygon.) Then P' is the polygon required. q. e. f. Proof. P':P = s"':s\ §334 ( The areas of two similar polygons are to each other as the squares on any two corresponding sides.) But s'^ : s'^ — n: m. Const. Therefore P^ : P = n : m, hj Ax. 8. q.e.d. This problem enables us to construct a square that is twice a given square or half a given square, to construct an equilateral triangle that shall be any number of times a given equilateral triangle, and in general to enlarge or to reduce any figure in a given ratio. An architect's drawl- ing, for example, might need to be enlarged so as to be double the area of the original, and the scale could be found by this method. EXERCISES 223 EXERCISE 55 Problems of Computation 1. The sides of a triangle are 0.7 in., 0.6 in., and 0.7 in. respectively. Is the largest angle acute, right, or obtuse ? 2. The sides of a triangle are 5.1 in., 6.8 in., and 8.5 in. respectively. Is the largest angle acute, right, or obtuse ? 3. Find the area of an isosceles triangle whose perimeter is 14 in. and base 4 in. (One decimal place.) 4. Find the area of an equilateral triangle whose perimeter is 18 in. (One decimal place.) 5. Find the area of a right triangle, the hypotenuse being 1.7 in. and one of the other sides being 0.8 in. 6. Find the ratio of the altitudes of two triangles of equal area, the base of one being 1.5 in. and that of the other 4.5 in. 7. The bases of a trapezoid are 34 in. and 30 in., and the altitude is 2 in. Find the side of a square having the same area. 8. What is the area of the isosceles right triangle in which the hypotenuse is V2 ? 9. AVhat is the area of the isosceles right triangle in which the hypotenuse is 7 V2 ? 10. If the side of an equilateral triangle is 2 V3, what is the altitude of the triangle ? the area of the triangle ? 11. If the side of an equilateral triangle is 1 ft., what is the area of the triangle ? 12. If the area of an equilateral triangle is 43.3 sq. in., what is the base of the triangle ? (Take V3 = 1.732.) 13. The sides of a triangle are 2.8 in., 3.5 in., and 2.1 in. respectively. Draw the figure carefully and see what kind of a triangle it is. Verify this conclusion by applying a geometric test, and find the area of the triangle. 224 BOOK lY. PLANE GEOMETRY EXERCISE 56 Theorems 1. The area of a rhombus is equal to half the product of its diagonals. 2. Two triangles are equivalent if the base of the first is equal to half the altitude of the second, and the altitude of the first is equal to twice the base of the second. 3. The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle. 4. Two parallelograms are equivalent if their altitudes are reciprocally proportional to their bases. 5. If equilateral triangles are constructed on the sides of a right triangle, the triangle on the hypotenuse is equivalent to the sum of the triangles on the other two sides. 6. If similar polygons are constructed on the sides of a right triangle, as corresponding sides, the polygon on the hypotenuse is equivalent to the sum of the polygons on the other two sides. Ex. 6 is one of the general forms of the Pythagorean Theorem. 7. If lines are drawn from any point within a parallelogram to the four vertices, the sum of either pair of triangles with parallel bases is equivalent to the sum of the other pair. 8. Every line drawn through the intersection of the diag- onals of a parallelogram bisects the parallelogram. 9. The line that bisects the bases of a trapezoid divides the trapezoid into two equivalent parts. 10. If either diagonal of a trapezoid bisects it, the trapezoid is a parallelogram. 11. The triangle formed by two lines drawn from the mid- point of either of the nonparallel sides of a trapezoid to the opposite vertices is equivalent to half the trapezoid. EXERCISES 225 EXERCISE 57 Problems of Construction 1. Given a square, to construct a square of half its area. 2. To construct a right triangle equivalent to a given oblique triangle. 3. To construct a triangle equivalent to the sum of two given triangles. 4. To construct a triangle equivalent to a given triangle, and having one side equal to a given line. 5. To construct a rectangle equivalent to a given parallelo- gram, and having its altitude equal to a given line. 6. To construct a right triangle equivalent to a given tri- angle, and having one of the sides of the right angle equal to a given line, 7. To construct a right triangle equivalent to a given tri- angle, and having its hypotenuse equal to a given line. 8. To divide a given triangle into two equivalent parts by a line through a given point P in the base. 9. To draw from a given point P in the base ylJ5 of a tri- angle ABC 2^ line to AC produced, so that it may be bisected by BC. 10. To find a point within a given triangle such that the lines from this point to the vertices shall divide the triangle into three equivalent triangles. 11. To divide a given triangle into two equivalent parts by a line parallel to one of the sides. 12. Through a given point to draw a line so that the seg- ments intercepted between the point and perpendiculars drawn to the line from two other given points may have a given ratio. 13. To find a point such that the perpendiculars from it to the sides of a given triangle shall be in the ratio p, q, r. 226 BOOK IV. PLANE GEOMETRY EXERCISE 58 Review Questions 1. What is meant by the area of a surface ? Illustrate. 2. What is the difference between equivalent figures and congruent figures ? 3. State two propositions relating to the ratio of one rectangle to another. 4. Given the base and altitude of a rectangle, how is the area found ? Given the area and base, how is the altitude found ? 5. How do you justify the expression, " the product of two lines " ? " the quotient of an area by a line " ? 6. Can a triangle with a perimeter of 10 in. have the same area as one with a perimeter of 1 in. ? Is the same answer true for two squares ? 7. Can a parallelogram with a perimeter of 10 in. have the same area as a rectangle with a perimeter of 1 in. ? Is the same answer true for two rectangles ? 8. Explain how the area of an irregular field with straight sides may be found by the use of the theorems of Book IV. 9. A triangle has two sides 5 and 6, including an angle of 70°, and another triangle has two sides 2 and 7^, including an angle of 70°. What is the ratio of the areas of the triangles ? - 10. Two similar triangles have two corresponding sides 5 in. and 15 in. respectively. The larger triangle has how many times the area of the smaller ? 11. Given the hypotenuse of an isosceles right triangle, how do you proceed to find the area ? 12. Given three sides of a triangle, what test can you apply to determine whether or not it is a right triangle ? 13. Suppose you wish to construct a square equivalent to a given polygon, how do you proceed ? BOOK V REGULAR POLYGONS AND CIRCLES 357. Regular Polygon. A polygon that is both equiangular and equilateral is called a regular polygon. Familiar examples of regular polygons are the equilateral triangle and the square. It is proved in Prop. I (§ 362) that a circle may be circumscribed about, and a circle may be inscribed in, any regular polygon, and that these circles are concentric (§ 188) . 358. Radius. The radius of the circle cir- cumscribed about a regular polygon is called the radius of the polygon. In this figure r is the radius of the polygon. 359. Apothem. The radius of the circle inscribed in a regular polygon is called the apothem of the polygon. In the figure a is the apothem of the polygon. The apothem is evi- dently perpendicular to the side of the regular polygon (§ 185) . 360. Center. The common center of the circles circumscribed about and inscribed in a regular polygon is called the center of the polygon. In the figure O is the center of the polygon. 361. Angle at the Center. The angle between the radii drawn to the extremities of any side of a regular polygon is called the angle at the center of the polygon. In the figure m is the angle at the center of the polygon. It is evidently subtended by the chord which is the side of the inscribed polygon. 227 228 BOOK V. PLANE GEOMETRY Proposition I. Theorem 362. A circle may he circumscribed about, and a circle may he inscribed in, any regular polygon. Given the regular polygon ABCDE. To prove that 1. a circle may he circumscribed about ABCDE ; 2. a circle may be inscribed in ABCDE. Proof. 1. Let O be the center of the circle which may be passed through the three vertices A, B, and C. § 190 Draw OA, OB, OC, OD. Then OB = OC, §162 and AB = CD. §357 Furthermore Z CBA = Z DCB, i 357 and Z CBO = Z OCB. § 74 .\Z0BA = ZDC0. Ax. 2 .-.A OAB is congruent to AOCD. § 68 .'.OA=OD. §67 Therefore the circle that passes through A, B, C, passes also through D. In like manner it may be proved that the circle that passes through B, C, and D passes also through E; and so on. Therefore the circle described with as a center and OA as a radius will be circumscribed about the polygon, by § 205. q. e. d. REGULAR POLYGOKS AND CIRCLES 229 Proof. 2. Let O be the center of the circumscribed circle. D Since the sides of the regular polygon are equal chords of the circumscribed circle, they are equally distant from the center. § 178 Therefore the circle described with as a center, with the perpendicular from O to a side of the polygon as a radius, will be inscribed in the polygon, by § 205. q.e.d. 363. Corollary 1. The radius drawn to any vertex of a regular polygon bisects the angle at the vertex. 364. Corollary 2. The angles at the center of any regular polygon are equal, and each is supplementary to an interior angle of the polygon. For the angles at the center are corresponding angles of congruent triangles. If M is the mid-point of AB^ then since the A MOB and OBM are complementary what can we say of their doubles, AOB and CBA ? 365. Corollary 3. An equilateral polygon inscribed in a circle is a regular polygon. Why are the angles also equal ? 366. Corollary 4. An equiangular polygon circumscribed about a circle is a regular polygon. By joining consecutive points of contact of the sides of the polygon can you show that certain isosceles triangles are congruent, and thus prove the polygon equilateral ? 230 BOOK V. PLANE GEOMETRY Proposition II. Theorem 367. If a circle is divided into any number of equal arcs, the chords joining the successive points of division form a regular inscribed polygon; and the tangents drawn at the points of division form a regidar circum- scribed polygon. Given a circle divided into equal arcs by ^, 5, C, Z), and E^ AB^ BCj CD, DE, and EA being chords, and PQ, QR, RS, ST, and TP being tangents at B, C, D, E, and A respectively. To prove that 1. ABODE is a regular polygon ; 2. PQRST is a regular polygon. Proof. 1. Since the arcs are equal by construction, .-. AB = BC = CD = DE = EA. § 170 .'. ABODE is a regular polygon. § 365 {An equilateral polygon inscribed in a circle is a regular polygon.) Proof. 2. AP = Z.Q = ZR = ZS = ZT. §221 {An Z formed by two tangents is measured by half the difference of the intercepted arcs.) .*. PQRST is a regular polygon. § 366 {An equiangular polygon circumscribed about a circle is a regular polygon.) Q. E. D. KEGULAR POLYGONS AND CIRCLES 231 368. Corollary 1. Taiigents to a circle at the vertices of a regular inscribed polygon form a regular circumscribed poly- gon of the same number of sides. 369. Corollary 2. Tangents to a circle at the mid-points of the arcs subtended by the sides of a regular inscribed polygon form a regular circumscribed polygon^ whose sides are parallel to the sides of the inscribed polygon and whose vertices lie on the radii (^produced^ of ^^ the inscribed polygon. For two corresponding sides, AB and A'B'^ are perpendicular to OM (§§ 176, 185), and are parallel (§ 95) ; and the tangents MB' and iVJB', intersecting at a point equidistant from OM and ON (§ 192), intersect upon the bisector of the Z MON {% 152) ; that is, upon the radius OB (§ 363). 370. Corollary 3. Lines draum from each vertex of a regular polygon to the 7nid-points of the adjacent arcs subtended by the sides of the polygon form a regular inscribed polygon of double the number of sides. 371. Corollary 4. Tangents at the mid- points of the arcs between adjacent points of contact of the sides of a regular circumscribed polygon form a regular circumscribed polygon ^^ of double the number of sides. j 372. Corollary 5. The perimeter of a regular inscribed polygon is less than that of a regular inscribed polygon of double the number of sides; and the perimeter of a regular circumscribed polygori is greater than that of a regular cir- cumscribed polygofi of double the number of sides. 232 BOOK V. PLANE GEOMETRY EXERCISE 59 1. Find the radius of the square whose side is 5 in. 2. Find the side of the square whose radius is 7 in. 3. Find the radius of the equilateral triangle whose side is 2 in. ^ 4. Find the side of the equilateral triangle whose radius is 3 in. 5. Find the apothem of the equilateral triangle whose side is VS in. 6. Find the side of the equilateral triangle whose apothem is 2 V3 in. 7. How many degrees are there in the angle at the center of an equiangular triangle ? of a regular hexagon ? 8. Given an equilateral triangle inscribed in a circle, to circumscribe an equilateral triangle about the circle. 9. Given an equilateral triangle inscribed in a circle, to in- scribe a regular hexagon in the circle, and to circumscribe a regular hexagon about the circle. 10. Given a square inscribed in a circle, to inscribe a regular octagon in the circle, and to circumscribe a regular octagon about the circle. 11. How many degrees are there in the angle at the center of a regular octagon ? in each angle of a regular octagon ? in the sum of these two angles ? 12. What is the area of the square inscribed in a circle of radius 2 in.? 13. The apothem of an equilateral triangle is equal to half the radius. 14. Prove that the apothem of an equilateral triangle is equal to one fourth the diameter of the circumscribed circle. From this show how an equilateral triangle may be inscribed in a circle. REGULAE POLYGONS AND CIRCLES Proposition III. Theorem 233 373. Tivo regular polygons of the same numher of sides are similar. §145 Given the regular polygons P and P', each having n sides. To prove that P and P' are similar. 2(71-2) Proof. Each angle of either polygon = rt. A. 2 (n — 2) {Each Z of a regular polygon of n sides is equal to — rt. A.) Hence the polygons P and P' are mutually equiangular. Furthermore, '.' AB = BC = CD = DE = EA, and A'B' = B'C = CD' = D'E' = E'A', § 357 AB _ BC _ CD _ DE _ EA ' ' A'B'~ B'C'~ CD' ~ D'E' ~ E'A'' Hence the polygons have their corresponding sides propor- tional and their corresponding angles equal. Therefore the two polygons are similar, by § 282. q.e.d. 374. Corollary. The areas of two regular polygons of the same numher of sides are to each other as the squares on any two corresponding sides. Por the areas of two similar polygons are to each other as tlie squares on any two corresponding sides (§ 334), and two regular polygons of the same number of sides are similar. Ax. 4 234 BOOK V. PLANE GEOMETEY Proposition IV. Theorem 375. The perimeters of two regular p)olygons of the same number of sides are to each other as their radii, and also as their apothems. A MB A' M' B' Given the regular polygons with perimeters p and p\ radii r and r', apothems a and a', and centers O and 0' respectively. To prove that p : p' = r : r' = a: a'. Proof. Since the polygons are similar, .\p:2)' = AB:A'B'. Furthermore in the isosceles AOAB and O'A'B', and OA : OB = O'A' : O'B'. {For each of these ratios equals 1.) .-.the AOAB and O'A'B' are similar. .'.AB:A'B' = r:r'. Also A A MO and A'M'O' are similar. .'. r :r' = a : a'. .-. j9 i]^' = r : r' = a : a', by Ax. 8. 376. Corollary. The areas of two regular polygons of the same number of sides are to each other as the squares on the radii of the circumscribed circles, and also as the squares on the radii of the inscribed circles. §373 §291 §364 Ax. 8 §288 §282 §286 §282 Q. E. D. KEGULAR POLYGONS AND CIRCLES 235 EXERCISE 60 1. Eind the ratio of the perimeters and the ratio of the areas of two regular hexagons, their sides being 2 in. and 4 in. respectively. 2. Find the ratio of the perimeters and the ratio of the areas of two regular octagons, their sides having the ratio 2 : 6. 3. Eind the ratio of the perimeters of two squares whose areas are 121 sq. in. and 30i sq. in. respectively. 4. Eind the ratio of the perimeters and the ratio of the areas of two equilateral triangles whose altitudes are 3 in. and 12 in. respectively. 5. The area of one equiangular triangle is nine times that of another. Required the ratio of their altitudes. 6. The area of the cross section of a steel beam 1 in. thick is 12 sq. in. What is the area of the cross section of a beam of the same proportions and 1^ in. thick ? 7. Squares are inscribed in two circles of radii 2 in. and 6 in. respectively. Eind the ratio of the areas of the squares, and also the ratio of the perimeters. 8. Squares are inscribed in two circles of radii 2 in. and 8 in. respectively, and on the sides of these squares equi- lateral triangles are constructed. What is the ratio of the areas of these triangles ? 9. A round log a foot in diameter is sawed so as to have the cross section the largest square possible. What is the area of this square ? What would be the area of the cross section of the square beam cut from a log of half this diameter ? 10. Every equiangular polygon inscribed in a circle is regular if it has an odd number of sides. 11. Every equilateral polygon circumscribed about a circle is regular if it has an odd number of sides. 236 BOOK V. PLANE GEOMETRY Proposition V. Theorem 377. If the number of sides of a regular inscribed polygon is indefinitely increased, the ai^otheni of the polygon approaches the radius of the circle as its limit. Given a regular polygon of n sides inscribed in the circle of radius OA^ s being one side and a the apothem. To prove that a approaches r as a limits if n is increased indefinitely. Proof. We know that a<r. § 86 Then since r~a<AM, § 112 and AM<s, §174 .\r — a<,s. Ax. 10 If n is taken sufficiently great, s, and consequently r— a, can be made less than any assigned positive value, however small. Since r—a can become and remain less than any assigned positive value by increasing n, it follows that r is the limit of a, by § 204. q.'e.d. 378. Corollary. // the number of sides of a regular in- scribed polygon is indefinitely increased^ the square on the apothem approaches the square on the radius of the circle as its limit. For r2 - a2 ^ AM'^. § 338 Therefore by taking n sufficiently great, s, and consequently AM, and consequently r^ — a^, approaches zero as its limit. REGULAR POLYGONS AND CIRCLES 237 Proposition VI. Theorem 379. An arc of a circle is less than a line of any kind that envelops it on the convex side and has the same extremities. Given BCA an arc of a circle, AB being its chord. To prove that the arc BCA is less than a line of any hind that envelops this arc and terminates at A and B. Proof. Of all the lines that can be drawn, each to include the area ABC between itself and the chord AB, there must be at least one shortest line ; for all the lines are not equal. Let BDA be any kind of line enveloping BCA as stated. The enveloping line BDA cannot be the shortest ; for draw- ing ECF tangent to the arc BCA at any point C, the line BFCEA<BFDEA, since FCE< FDE. Post. 3 In like manner it can be shown that no other enveloping line can be the shortest. .*. BCA is shorter than any enveloping line. q.e.d. 380. Corollary. A circle is less than the perimeter of any polygon circumscribed about it. 381. Circle as a Limit. From Prop. VI we may now assume : 1. The circle is the limit which the perimeters of regular in- scribed polygons and of similar circumscribed polygons approach, if the number of sides of the polygons is indefinitely increased. 2. The area of the circle is the limit which the areas of the inscribed and circumscribed polygons approach. 238 BOOK V. PLANE GEOMETEY Proposition YII. Theorem 382. Tico circumferences hcwe the same ratio as their radii. Given the circles with circumferences c and c', and radii r and r' respectively. To prove that c: c' = r : r'. Proof. Inscribe in the circles two similar regular polygons, and denote their perimeters by ^ and^>'. Then p:2)' = r:7''. §375 Conceive the number of sides of these regular polygons to be indefinitely increased, the polygons continuing similar. Thenp and 2^' will have c and c' as limits. § 381 But the ratio p :p' will always be equal to the ratio ?• : r' .'. jyr' =2)'t'. .'. cr' = c'r. ,'.c:c' = r:r', by §264. 383. Corollary 1. The ratio of any circle to its diameter is constant. Why does c : c'= 2 r : 2 r' ? Then why does c : 2 r = c' : 2 r' ? 384. The Symbol tt. The constant ratio of a circle to its diameter is represented by the Greek letter rrr (pi). 385. Corollary 2. In any circle c=2irr. §375 §261 §207 Q.E.D. For, by definition, 17 = 2r REGULAR POLYGONS AND CIRCLES 239 Proposition VIII. Theorem 386. The area of a regular polygon is equal to half the product of its apothem hy its perimeter. A M B Given the regular polygon ABCDEF, with apothem a, perimeter />, and area s. To prove that * — 2 ^P' Proof. Draw the radii OA, OB, OC, etc., to the successive vertices of the polygon. The polygon is then divided into as many triangles as it has sides. The apothem is the common altitude of these A, and the area of each A is equal to \a multiplied by the base. § 325 Hence the area of all the triangles is equal to \ a multiplied by the sum of all the bases. Ax. 1 But the sum of the areas of all the triangles is equal to the area of the polygon. Ax. 11 And the sum of all the bases of the triangles is equal to the perimeter of the polygon. Ax. 11 .-. s = ^ ap. Q.E. D. 387. Similar Parts. In different circles similar arcs, similar sectors, and similar segments are such arcs, sectors, and seg- ments as correspond to equal angles at the center. For example, two arcs of 30° in different circles are similar arcs, and the corresponding sectors are similar sectors. 240 BOOK V. PLANE GEOMETRY Proposition IX. Theorem 388. The area of a circle is equal to half the product of its radius hy its circumference. Given a circle with radius r, circumference c, and area s. To prove that s = 1 re. Proof. Circumscribe any regular polygon of n sides, and denote the perimete*- of this polygon by p and its area by s\ Then since r is its apothem, s' = ^ 77^. § 386 Conceive n to be indefinitely increased. Then since p approaches c as its limit, § 381 and r is constant, -'■ \rp approaches \rc as its limit. Also s' approaches s as its limit. § 381 But s' = \rp always. § 386 .\s = lrc, by § 207. q.e.d. 389. Corollary 1. The area of a circle is equal to ir times the square on its radius. For the area of the O = \rc = \r x ^irr — tit^. 390. Corollary 2. The areas of two circles are to each other as the squares on their radii. 391. Corollary 3. The area of a sector is equal to half the product of its radius hy its are. area of sector arc of sector For area of circle circle EEGULAR POLYGONS AND CIRCLES 241 EXERCISE 61 1. Two circles are constructed with radii 1^^ in. and 4^ in. respectively. The circumference of the second is how many times that of the first? 2. The circumference of one circle is three times that of another. The square on the radius of the first is how many times the square on the radius of the second? 3. The circumference of one circle is 2^ times that of another. The equilateral triangle constructed on the diameter of the first has how many times the area of the equilateral triangle constructed on the diameter of the second ? 4. A circle with a diameter of 5 in. has a circumference of 15.708 in. What is the circumference of a pipe that has a diameter of 2 in. ? 5. A wheel with a circumference of 4 ft. has a diameter of 1.27 ft., expressed to two decimal places. What is the cir- cumference of a wheel with a diameter of 1.58| ft. ? 6. A regular hexagon is 2 in. on a side. Find its apothem and its area to two decimal .places. 7. An equilateral triangl^is 2 in. on a side. Find its apothem and its area to two decimal places. 8. The radius of one circle is 2^ times that of another. The area of the smaller is 15.2 sq. in. What is the area of the larger ? 9. The radius of one circle is 3^ times that of another. The area of the smaller is 17.75 sq. in. What is the area of the larger ? 10. The circumferences of two cylindrical steel shafts are respectively 3 in. and 1^ in. The area of a cross section of the first is how many times that of a cross section of the second ? 11. The arc of a sector of a circle 2i in. in diameter is 1| in. What is the area of the sector ? 242 BOOK V. PLANE GEOMETRY Proposition X. Problem 392. To inscribe a square in a given circle. Given a circle with center O. Required to inscribe a square in the given circle. Construction. Draw two diameters A C and BD perpendicular to each other. § 228 Draw AB, BC, CD, and DA. Then A BCD is the square required. Proof. The A CBA, DCB, ADC, BAD are rt. A. {An Z inscribed in a semicircle is a rt. Z.) The A at the centef being rt. A, the arcs AB, BC, CD, and DA are equal, and the sides AB, BC, CD, and DA are equal. Hence the quadrilateral A BCD is a square, by § 65. q.e.d. 393. Corollary. To inscribe regular polygons of ^,1Q,Z2, 64, etc., sides in a given circle. By bisecting the arcs AB., BC, etc., a regular polygon of how many sides may be inscribed in the circle ? By continuing the process regular polygons of how many sides may be inscribed ? In general we may say that this corollary allows us to inscribe a reg- ular polygon of 2« sides, where n is any positive integer. As a special case it is interesting to note that n may equal 1. Q.E.F. §215 Const. §212 §170 PEOBLEMS OF CONSTRUCTION 243 Proposition XI. Problem 394. To inscribe a regular hexagon in a given circle. A>-~ — -^B Given a circle with center 0. Required to inscribe a regular hexagon in the given circle. Construction. From the center draw any radius, as OC. With C as a center, and a radius equal to OC, describe an arc intersecting the circle at D. Draw OD and CD. Then CD is a side of the regular hexagon required, and therefore the hexagon may be inscribed by applying CD six times as a chord. q.e.f. Proof. The A OCD is equiangular. § 75 {An equilateral triangle is equiangular.) Hence the Z COD is ^ of 2 rt. Zs, or ^ of 4 rt. A. § 107 .'. the arc CD is ^ of the circle. .'. the chord CD is a side of a regular inscribed hexagon, q. e. d. 395. Corollary 1. To inscribe an equilateral triangle in a given circle. By joining the alternate vertices of a regular inscribed hexagon, an equilateral triangle may be inscribed. 396. Corollary 2. To inscribe regular polygons of 12, 24, 48, etc., sides in a given circle. 244 BOOK V. PLANE GEOMETRY Proposition XII. Problem 397. To inscribe a regular decagon in a given circle. Given a circle with center O. Required to inscribe a regular decagon in the given circle. Construction. Draw any radius OA, and divide it in extreme and mean ratio, so that OA:OP=OP:AP. § 311 From A as a center, with a radius equal to OP, describe an arc intersecting the circle at B. Draw AB. Then A Bis a. side of the regular decagon required, and there- fore the regular decagon may be inscribed by applying AB ten Q. E. F. Draw PB and OB. OA : OP = OP : APj AB = OP. '.OA:AB = AB:AP. ZBAO = ZBAP. Hence the AOAB and BAP are similar. But the AOAB is isosceles. .*. A BAP, which is similar to AOAB, is isosceles, '§ 282 and AB = BP = OP. §62 times as a chord Proof. Then and Moreover, Const. Const. Ax. 9 Iden. §288 §162 PROBLEMS OF CONSTRUCTION .245 The A PBO being isosceles, the Z O = Z OBP. § 74 But the Z^P5 = Z0-hZ0^P = 2Z0. §111 Hence ABAP = 2/.0, and ^ ZOi5^ = 2ZO. Ax. 9 .-.the sum of the A of the A OAB = 5 Z O = 2 rt. Z, § 107 and Z O = 1 of 2 rt. A, or -^^ of 4 rt. A. Ax. 4 Therefore the arc ^jB is j^^ of the circle. § 212 .'. the chord ^5 is a side of a regular inscribed decagon. q.e.d. 398. Corollary 1. To inscribe a regular pentagon in a given circle. 399. Corollary 2. To inscribe regular polygons of 20, 40, 80, etc.^ sides in a given circle. By bisecting the arcs subtended by the sides of a regular inscribed decagon a regular polygon of how many sides may be inscribed ? By con- tinuing the process regular polygons of how many sides may be inscribed ? EXERCISE 62 If r denotes the radius of a regular inscribed polygon, a the apothem, s one side, A an angle, and C the angle at the center, show that: 1. In a regular inscribed triangle s = r Vs, a = ^r, A = 60°, C=120°. 2. In a regular inscribed quadrilateral s = r V2, a = ^r V2, A = 90°, C = 90°. 3. In a regular inscribed hexagon s = r,a= ^r Vs, A = 120°, C = 60°. 4. In a regular inscribed decagon s = ^^ 5-1) ^ ^ ^ 1^ VlO + 2 V5, ^ =144°, C= 36°. 246 BOOK V. PLANE GEOMETRY Proposition XIII. Problem 400. To inscribe in a given circle a recjulcir jpentadec- agon, or polygon of fifteen sides. Given a circle. Required to inscribe a regular pentadecagon in the given circle. Construction. Draw a chord PH equal to the radius of the circle, a chord PA equal to a side of the regular inscribed decagon, and draw AB. Then ^5 is a side of the regular pentadecagon required, and therefore the regular pentadecagon may be inscribed by apply- ing AB fifteen times as a chord. q.e.f. Proof. The arc PB is i of the circle, § 394 and the arc PA is ^^ of the circle. § 397 Hence the arc AB \?> ^ — J^j, or ^^, of the circle. Ax. 2 .*. the chord .45 is a side of the regular inscribed pentadec- agon required. q.e.d. 401. Corollary. To inscribe regular polygons (^f 30, 60, 120, etc.^ sides in a given circle. By bisecting the arcs AB^ BC, etc., a regular polygon of how many sides may be inscribed ? By continuing the process regular polygons of how many sides may be inscribed ? In general we may say that a regu- lar polygon of 15 • 2« sides may be inscribed in this manner. PKOBLEMS OF CONSTRUCTION 247 EXERCISE 63 1. A five-cent piece is placed on the table. How many five- cent pieces can be placed around it, each tangent to it and tangent to two of the others ? Prove it. 2. What is the perimeter of an equilateral triangle inscribed in a circle with radius 1 in. ? 3. What is the perimeter of an equilateral triangle circum- scribed about a circle with radius 1 in. ? 4. What is the perimeter of a regular hexagon circumscribed about a circle with radius 1 in, ? Required to circumscribe about a given circle the following regular polygons : 5. Triangle. 7. Hexagon. 9. Pentagon. 6. Quadrilateral. 8. Octagon. 10. Decagon. 11. Required to describe a circle whose circumference equals the sum of the circumferences of two circles of given radii. 12. Required to describe a circle whose area equals the sum of the areas of two circles of given radii. 13. Required to describe a circle having three times the area of a given circle. Required to construct an angle of : 14. 18°. 15. 36°. 16. 9°. 17. 12°. 18. 24°. Required to construct tvith a side of given length : 19. An equilateral triangle. 23. A regular pentagon. 20. A square. 24. A regular decagon. 21. A regular hexagon. 25. A regular dodecagon. 22. A regular octagon. 26. A regular pentadecagon. 27. From a circular log 16 in. in diameter a builder wishes to cut a column with its cross section as large a regular octagon as possible. Find the length of each side. 248 BOOK V. PLANE GEOMETRY Proposition XIV. Problem 402. Given the side and the radius of a regular in- scribed polygon, to find the side of the regular inscribed polygon of double the nuviber of sides. Given AB, the side of a regular inscribed polygon of radius OA. Required to find AP, a side of the regular inscribed poly- yon of double the number of sides. Solution. Denote the radius by r, and the side AB by s. Draw the diameter PQ 1. to AB, and draw AO and AQ. Then AM=\s. In the rt. A. 4 03/, oFf = 7^-\s^. Therefore OM = ^r - i s\ Since PM-^OM=r, therefore PM =r— OM Furthermore = r — V/*^ — 1 s^. AP =PQX PM. But PQ = 2r, and PM =r — Vr^ s\ .'. AP = 2 r(r - V?-" - i s% .'.AP = Vr(2r- V4r^-6-'). §174 §338 Ax. 5 Ax. 11 Ax. 2 Ax. 9 §298 Ax. 9 Ax. 5 Q. E. F. 403. Corollary. 7/ r =1, ^P = \/2 - V4 - s' PROBLEMS OF COMPUTATION Proposition XV. Problem 249 404. To find the numerical value of the ratio of the circmnference of a circle to its diameter. Given a circle of circumference c and diameter d. c Required to find the numerical value of - or ir. ct Solution. By § 385, 2 7rr = c. . • , tt = ^ c when r = 1. Let Sg (read " s sub six ") be the length of a side of a regular polygon of 6 sides, s^^ of 12 sides, and so on. If r = 1, by § 394, ^<?g = 1, and by § 403 we have Form of Computation = \/2-V4-l^ s^ = V2 - V4^0.51763809)2 s^3 = V2 - V4^ (0.26105238^ s^ = V2 - Vr- (0.13080626)^ ^m = V2 - V4 - (0.06543817)^ s^^ = V2- V4 - (0.03272346)2 s_g3 = V2 - V4^(0.01636228)^ .'.c = 6.28317 nearly ; that is^ TT is an incommensurable number. We generally take TT = 3.1416, or ^, and - = 0.31831. Length of Side Length of Perimeter 0.51763809 6.21165708 0.26105238 6.26525722 0.13080626 6.27870041 0.06543817 6.28206396 0.03272346 6.28290510 0.01636228 6.28311544 0.00818121 6.28316941 = 3.14159 nearly. Q.e.f. 250 BOOK V. PLAKE GEOMETRY EXERCISE 64 Problems of Computation Using the value S.1416 for tt, find the cireumferences of circles with radii as follows : 1. 3 in. 3. 2.7 in. 5. 7| in. 7. 2 ft. 8 in. 2. 5 in. 4. 3.4 in. 6. 6| in. 8. 3 ft. 7 in. Find the circumferences of circles with diameters as follows : 9. 9 in. 11. 5.9 in. 13. 2^ ft. 15. 29 centimeters. 10. 12 in. 12. 7.3 in. 14. 3^ in. 16. 47 millimeters. Find the radii of circles with circumferences as follows : 17. Tit. 19. 15.708 in. 21. 18.8496 in. 23. 345.576 ft. 18. 3^77. 20. 21.9912 in. 22. 125.664 in. 24. 3487.176 in. Find the diameters of circles with circumferences as follows : 25. 15 TT. 27. 2 7rr. 29. 188.496 in. 31. 3361.512 in. 26. 7r\ 28. 7 7ral 30. 219.912 in. 32. 3173.016 in. Find the areas of circles ivith radii as folloivs : 33. 5 X. 35. 27 ft. 37. 3^ in. 39. 2 ft. 6 in. 34. 2 TT. 36. 4.8 ft. 38. 4f in. 40. 7 ft. 9 in. Find the areas of circles with diameters as follows : 41. 16aZ». 43.2.5 ft. 45. 3| yd. 47. 3 ft. 2 in. 42. 24 tt". 44. 7.3 in. 46. 4f yd. 48. 4 ft. 1 in. Find the areas of circles tvith circumferences as folloivs : 49. 2 7r. 51. ira. 53. 18.8496 in. 55. 333.0096 in. 50. 4 7r. 52. 14 Tra^. 54. 329.868 in. 56. 364.4256 in. Find the radii of circles with areas as follows : 57. -n-a^*. 59. tt. 61. 12.5664. 63. 78.54. 58. 4 7rmV. 60. 2 7r. 62. 28.2744. 64. 113.0976. EXEKCISES 251 EXERCISE 65 Problems of Construction 1. To inscribe in a given circle a regular polygon similar to a given regular polygon. 2. To divide by a concentric circle the area of a given circle into two equivalent parts. 3. To divide by concentric circles the area of a given circle into n equivalent parts. 4. To describe a circle whose circumference is equal to the difference of two circumferences of given radii. 5. To describe a circle the ratio of whose area to that of a given circle shall be equal to the given ratio m, : n. 6. To construct a regular pentagon, given one of the diagonals. 7. To draw a tangent to a given circle such that the seg- ment intercepted between the point of contact and a given line shall have a given length. 8. In a given equilateral triangle to inscribe three equal circles tangent each to the other two, each circle being tangent to two sides of the triangle. 9. In a given square to inscribe four equal circles, so that each circle shall be tangent to two of the others and also tangent to two sides of the square. 10. In a given square to inscribe four equal circles, so that each circle shall be tangent to two of the others and also tangent to one side and only one side of the square. 11. To draw a common secant to two given circles exterior to each other, such that the intercepted chords shall have the given lengths a and b. 12. To draw through a point of intersection of two given intersecting circles a common secant of a given length. 252 BOOK V. PLANE GEOMETRY EXERCISE 66 Problems of Loci 1. Find the locus of the center of the circle inscribed in a triangle that has a given base and a given angle at the vertex. 2. Find the locus of the intersection of the perpendiculars from the three vertices to the opposite sides of a triangle that has a given base and a given angle at the vertex. 3. Find the locus of the extremity of a tangent to a given circle, if the length of the tangent is equal to a given line. 4. Find the locus of a point from which tangents drawn to a given circle form a given angle. 5. Find the locus of the mid-point of a line drawn from a given point to a given line. 6. Find the locus of the vertex of a triangle that has a given base and a given altitude. 7. Find the locus of a point the sum of whose distances from two given parallel lines is constant. 8. Find the locus of a point the difference of whose dis- tances from two given parallel lines is constant. 9. Find the locus of a point the sum of whose distances from two given intersecting lines is constant. 10. Find the locus of a point the difference of whose dis- tances from two given intersecting lines is constant. 11. Find the locus of a point whose distances from two given points are in the ratio m : n. 12. Find the locus of a point whose distances from two given parallel lines are in the ratio m : n. 13. Find the locus of a point whose distances from two given intersecting lines are in the ratio m:n. 14. Find the locus of a point the sum of the squares of whose distances from two given points is constant. EXERCISES 253 EXERCISE 67 Examination Questions 1. Each side of a triangle is 2 ti centimeters, and about each vertex as a center, with a radius of n centimeters, a circle is described. Find the area bounded by the three arcs that lie outside the trid:ngle, and the area bounded by the three arcs that lie within the triangle. 2. Upon a line AB 2i segment of a circle containing 240° is constructed, and in the segment any chord CD subtending an arc of 60° is drawn. Find the locus of the intersection of ^C and J5Z), and also of the intersection oi AD and BC. 3. Three successive vertices of a regular octagon are A, B, and C. If the length AB is a, compute the length A C. 4. The areas of similar segments of circles are proportional to the squares on their radii. 5. An arc of a certain circle is 100 ft. long and subtends an angle of 25° at the center. Compute the radius of the circle correct to one decimal place. 6. Given a circle whose radius is 16, find the perimeter and the area of the regular inscribed octagon. 7. If two circles intersect at the points A and B, and through A a variable secant is drawn, cutting the circles in C and D, the angle CBD is constant for all positions of the secant. 8. If A and B are two fixed points on a given circle, and P and Q are the extremities of a variable diameter of the same circle, find the locus of the point of intersection of the lines AP and BQ. 9. The radius of a circle is 10 ft. Two parallel chords are drawn, each equal to the radius. Find that part of the area of the circle lying between the parallel chords. The propositions in Exercise 67 are taken from recent college entrance examination papers. 254 BOOK V. PLANE GEOMETRY EXERCISE 68 Formulas If r denotes the radius of a circle^ and s one side of a reg- ular inscribed polygon, prove the following, and find the value of s to two decimal places when r —1 : 1. In an equilateral triangle s = r VS. 2. In a square s = r V2. 3. In a regular pentagon 5 = ^ ?• V 10 — 2 V 5. 4. In a regular hexagon s = r. 5. In a regular octagon s = r v2 — V2. 6. In a regular decagon s = ^ r (Vs — 1). 7. In a regular dodecagon s = r\2 — V 3. 8. A regular pentagon is inscribed in a circle whose radius is r. If the side is s, find the apothem. 9. A regular polygon is inscribed in a circle whose radius is r. If the side is s, show that the apothem is \ V4 7^ — s\ 10. If the radius of a circle is r, and the side of an inscribed regular polygon is s, show that the side of the similar cir- cumscribed regular polygon is 11. Three equal circles are described, each tangent to the other two. If the common radius is r, find the area contained between the circles. 12. Given p, P, the perimeters of regular polygons of n sides inscribed in and circumscribed about a given circle. Eind p', P', the perimeters of regular polygons of 2n sides inscribed in and circumscribed about the given circle. 13. A circular plot of land d ft. in diameter is surrounded by a walk w ft. wide. Find the area of the circular plot and the area of the walk. EXERCISES 255 EXERCISE 69 Applied Problems 1. The diameter of a bicycle wheel is 28 in. How many revolutions does the wheel make in going 10 mi. ? 2. Find the diameter of a carriage wheel that makes 264 revolutions in going half a mile. 3. A circular pond 100 yd. in diameter is surrounded by a walk 10 ft. wide. Find the area of the walk. 4. The span (chord) of a bridge in the form of a circular arc is 120 ft., and the highest point of the arch is 15 ft. above the piers. Find the radius of the arc. 5. Two branch water pipes lead into a main pipe. It is necessary that the cross-section area of the main pipe shall equal the sum of the cross sections of the two branch pipes. The diameters of the branch pipes are respectively 3 in. and 4 in. Required the diameter of the main pipe. 6. A kite is made as here shown, the semicircle / , \ having a radius of 9 in., and the triangle a height of 25 in. Find the area of the kite. 7. In making a drawing for an arch it is required to mark off on a circle drawn with a radius of 5 in. an arc that shall be 8 in. long. This is best done by finding the angle at the center. How many degrees are there in this angle ? 8. In an iron washer here shown, the diameter of the hole is 1| in. and the width of the washer is I in. Find the area of one face of the washer. 9. Find the area of a fan that opens out into a sector of 120°, the radius being 9| in. 10. The area of a fan that opens out into a sector of 111° is 96.866 sq. in. What is the radius ? (Take tt = 3.1416.) 256 BOOK V. PLANE GEOMETRY EXERCISE 70 Review Questions 1. What is meant by a regular polygon ? by its radius ? by its center? by its apothem ? 2. What other names are there for a regular triangle and a regular quadrilateral ? 3. If one angle of a regular polygon is known, how can the number of sides be determined? 4. The sides of two regular polygons of n sides are respec- tively s and s\ What is the ratio of their radii ? of their apothems ? of their perimeters ? of their areas ? 5. The diameters of two circles are d and d' respectively. What is the ratio of their radii ? of their circumferences ? of their areas ? 6. If the number of sides of a regular inscribed polygon is indefinitely increased, what is the limit of the apothem? of each side ? of the perimeter ? of the area ? of the angle at the center ? of each angle of the polygon ? 7. How do you find the area of a regular polygon ? of an irregular polygon ? of a square ? of a triangle ? of a parallelo- gram ? of a circle ? of a trapezoid ? of a sector ? 8. What regular polygons have you learned to inscribe in a circle ? Name three regular polygons that you have not learned to inscribe. 9. Given the circumference of a circle, how can the area of the circle be found ? 10. Given the area of a circle, how can the circumference of the circle be found ? 11. What is the radius of the circle of which the number of linear units of circumference is equal to the number of square units of area ? EXEECISES 257 EXERCISE 71 General Review of Plane Geometry Write a classification of the different kinds of: 1. Lines. 3. Triangles. 5. Polygons. 2. Angles. 4. Quadrilaterals. 6. Parallelograms. State the conditions under which : 7. Two triangles are congruent. 8. Two parallelograms are congruent. 9. Two triangles are similar. 10. Two straight lines are parallel. 11. Two parallelograms are equivalent. 12. Two polygons are similar. Complete the following statements in the most general manner: 13. In any triangle the square on the side opposite • • •. 14. If two parallel lines are cut by a transversal, • • •. 15. If four quantities are in proportion, they are in pro- portion by • • • . 16. If two secants of a circle intersect, the angle formed is measured by •••. 17. The perimeters of two similar polygons are to each other as • • • . 18. The areas of two similar polygons are to each other as • • •. 19. The area of a circle is equal to • • •. 20. In the same circle or in equal circles equal chords • • •. 21. In the same circle or in equal circles the central angles subtended by two arcs are • • • . 22. If two secants of a circle intersect within, on, or outside the circle, the product of • • • . 258 BOOK V. PLANE GEOMETRY 23. If four lines meet in a point so that the opposite angles are equal, these lines form two intersecting straight lines. 24. If squares are constructed outwardly on the six sides of a regular hexagon, the exterior vertices of these squares are the vertices of a regular dodecagon. 25. In a right triangle the line joining the vertex of the right angle to the mid-point of the hypotenuse is equal to half the hypotenuse. 26. No two lines drawn from the vertices of the base angles of a triangle to the opposite sides can bisect each other. 27. The rhombus is the only parallelogram that can be cir- cumscribed about a circle. 28. The square is the only rectangle that can be circum- scribed about a circle. 29. No oblique parallelogram can be inscribed in a circle. 30. If two triangles have equal bases and equal vertical angles, the two circumscribing circles have equal diameters. 31. If the inscribed and circumscribed circles of a triangle are concentric, the triangle is equilateral. 32. If the three points of contact of a circle inscribed in a tri- angle are joined, the angles of the resulting triangle are all acute. 33. The diagonals of a regular pentagon intersect at the vertices of another regular pentagon. 34. If two perpendicular radii of a circle are produced to intersect a tangent to the circle, the other tangents from the two points of intersection are parallel. 35. The line that joins the feet of the perpendiculars drawn from the extremities of the base of an isosceles triangle to the equal sides is parallel to the base. 36. The sum of the perpendiculars drawn to the sides of a regular polygon from any point within the polygon is equal to the apothem multiplied by the number of sides. EXERCISES 259 37. If two consecutive angles of a quadrilateral are right angles, the bisectors of the other two angles are perpendicular. 38. If two opposite angles of a quadrilateral are right angles, the bisectors of the other two angles are parallel. 39. The two lines that join the mid-points of opposite sides of a quadrilateral bisect each other. 40. The sum of the angles at the vertices of a five-pointed star is equal to two right angles. 41. The segments of any line intercepted between two con- centric circles are equal. 42. The diagonals of a trapezoid divide each other into segments which are proportional. 43. Given the mid-points of the sides of a triangle, to con- struct the triangle. 44. To divide a given triangle into two equivalent parts by a line through one of the vertices. 45. To draw a tangent to a given circle that shall also be perpendicular to a given line. 46. To divide a given line into two segments such that the square on one shall be double the square on the other. 47. If any two consecutive sides of an inscribed hexagon are respectively parallel to their opposite sides, the remaining two sides are parallel. 48. If through any given point in the common chord of two intersecting circles two other chords are drawn, one in each circle, their four extremities will all lie on a third circle. 49. If two chords intersect at right angles within a circle, the sum of the squares on their segments equals the square on the diameter. Investigate the case in which the chords intersect outside the circle ; also the case in which they inter- sect on the circle. 260 BOOK V. PLANE GEOMETRY 50. The lines bisecting any angle of an inscribed quadrilateral and the opposite exterior angle intersect on the circle. 51. The sum of the perpendiculars from any point in an equilateral triangle to the three sides is constant. 52. The perpendiculars from the vertices of a triangle upon the opposite sides cut one another into segments that are reciprocally proportional to each other. 53. The area of a triangle is equal to half the product of its perimeter by the radius of the inscribed circle. 54. The perimeter of a triangle is to one side as the perpen- dicular from the opposite vertex is to the radius of the inscribed circle. 55. The area of a square inscribed in a semicircle is equal to two fifths of the area of the square inscribed in the circle. 56. The diagonals of any inscribed quadrilateral divide it into two pairs of similar triangles. 57. To draw a line whose length is Vt^ in. 58. If two equivalent triangles are on the same base and the same side of the base, any line cutting the triangles, and par- allel to the base, cuts off equal areas from the triangles. 59. To divide a given arc of a circle into two parts such that their chords shall be in a given ratio. 60. The areas between two concentric circles may be found by multiplying half the sum of the two circumferences by the difference between the radii. 61. Find the length of the belt connecting two wheels of the same size, if the radius of each wheel is 18 in., the distance between the centers 6 ft., and 4 in. is allowed for sagging. 62. To construct a regular inscribed heptagon draftsmen sometimes use for a side half the side of an inscribed equi- lateral triangle. Construct such a figure with the compasses, and state whether the rule seems exact or only approximate. APPENDIX 405. Subjects Treated. Of the many additionar subjects that may occupy the attention of the student of plane geometry if time permits, two are of special interest. These are Symmetry, and Maxima and Minima. 406. Symmetric Points. Two points are said to be symmetric with respect to a p)oi7it, called the center of symmetry, if this third point bisects the straight line which joins the two points. Two points are said to be symmetric with respect to an axis, if a straight line, called the axis of symmetry, is the perpen- dicular bisector of the line joining them. 407. Symmetric Figure. A figure is said to be symmetric with resjject to a pjoint, if the point bisects every straight line drawn through it and terminated by the boundary of the figure. A figure is said to be symmetric with respect to an axis, if the axis bisects every /_ perpendicular through it and terminated by the boundary of the figure. Evidently this will be the case if one part coin- cides with another part when folded over the axis. 408. Two Symmetric Figures. Two figures are said to be symmetric ivith respect to a point or symmetric with respect to an axis, if every point of each has a correspond- ing symmetric point in the other. 261 B' 262 APPENDIX TO PLANE GEOMETRY Pkoposition I. Theorem A quadrilateral that has tioo adjacent sides and the other tivo sides equal, is symmetric loith respect to the diagonal joining the vertices of the angles formed hy the equal sides; and the diagonals are perpendicular to each other. 409 equal Given the quadrilateral ABCD, having AB equal to AD, and CB equal to CD, and having the diagonals AC and BD. To prove that the diagonal A C is an axis of symmetry^ and that AC is A. to BD. Proof. In the A yl/?C and ADC, AB = AD, and CB = CD, Given and AC = AC. Iden. .'. A ABC is congruent to A ADC. § 80 .•.ZBAC = ZCAD, and ZACB=:ZDCA. §67 Hence, if A ^5C is turned on ^C as an axis until it falls on A ADC, AB will fall on AD, CB on CD, and OB on OD. .-. the A ABC will coincide with the A ADC. .'.AC will bisect every perpendicular drawn through it and terminated by the boundary of the figure. .*. ^C is an axis of symmetry. § 407 .-. ^C is ± to BD, by § 406. Q.e.d. SYMMETRY 263 Proposition II. Theorem 410. If a figure is symmetric with respect to tioo axes j^erpendicular to each other, it is symmetric with respect to their intersection as a center. - C B L A E G Y' Given the figure ABCDEFGH, symmetric.with respect to the two perpendicular axes XX^, YY', which intersect at O. To prove that is the center of symmetry of the fiyure. Proof. Let P be any point in the perimeter. Draw PMQ J_ to YY\ and QNR _L to XX\ Then PQ is II to XX\ and QR is II to YY\ Draw PO, OR, and MN. Then QN = NR. {The figure is given as symmetric with respect to XX \) But QN = MO. .\NR = MO. .'. RO is equal and parallel to NM. In like manner, OP is equal and parallel to NM. .'. ROP is a straight line. .'.0 bisects PR, any straight line, and hence bisects every straight line drawn through and terminated by the perimeter. .*. O is the center of symmetry of the figure, by § 407. Q-E.d. §227 §95 §407 §127 Ax. 8 §130 §94 264 APPENDIX TO PLANE GEOMETRY EXERCISE 72 1. Draw a figure showing the number of axes of symmetry- possessed by a square. 2. Draw a figure showing the number of axes of symmetry possessed by a regular hexagon. 3. Draw a figure showing six of the unlimited number of axes of symmetry of a circle, and showing the center of symmetry. 4. Show by drawings that two congruent triangles may be placed in a position of symmetry with respect to an axis. In one of the drawings let a common side be the axis. 5. Show by a drawing that two congruent triangles may be placed in a position of symmetry with respect to a center. 6. Two figures symmetric with respect to an axis are con- gruent. 7. Two figures symmetric with respect to a center are con- gruent. 8. Make a list of quadrilaterals that are symmetric with respect to an axis. 9. Make a list of quadrilaterals that are symmetric with respect to a center. 10. What kinds of regular polygons are symmetric with re- spect both to a center and to an axis ? Prove this for the hexagon. 11. A circle is symmetric with respect to its center as a center of symmetry, and is also symmetric with respect to any diameter as an axis. 12. An isosceles triangle is symmetric with respect to an axis, and therefore the angles opposite the equal sides are equal. 13. Two tangents drawn to a circle from the same point are symmetric with respect to an axis. 14. The four common tangents to two given circles form, together with the circles, a figure symmetric with respect to the line of centers as an axis. MAXIMA AND MINIMA 265 411. Maxima and Minima. Among geometric magnitudes that satisfy given conditions, the greatest is called the maximum, and the smallest is called the minimum. The plural of maximum is maxima, and the plural of minimum is minima. Among geometric magnitudes that satisfy given conditions, there may be several equal magnitudes that are greater than any others. In this case all are called maxima. . Similarly there may be several minima magnitudes of a given kind. 412. Isoperimetric Polygons. Polygons which have equal perimeters are called isoperimetric polygons. If the circumference of a circle equals the perimeter of a polygon, the circle and the polygon are said to be isoperimetric, and similarly for all other closed figures in a plane. Proposition III. Theorem 413. Of all triangles having two given sides, that in which these sides include a right angle is the maximum. P A F B Given the triangles ABC and ABDy with AB and CA equal to AB and DA respectively, and with angle BAC a right angle. To prove that AABC> A ABD. Proof. Prom D draw the altitude DP. § 227 Then DA > DP. § 86 But DA = CA. Given .-. CA > DP. Ax. 9 .'.A ABC >AABD, by § 327. Q.e.d. 266 APPENDIX TO PLANE GEOMETRY Proposition IV. Theorem 414. Of all isoj^erimetric triangles having the same base the isosceles triangle is the maximum. B' yi ./' '/> / ^;^ /h" / 1 / -HQ Fig. 1 Fig. 2 §135 §215 §97 Given the triangles ABC and ABC having equal perimeters, and having AC equal to -BC, and AC not equal to BC\ To prove that A ABC > A ABC. Proof. Produce AC to B', making CB' = AC. Draw BB' and C'B', and draw C(2 II to AB. Then since AC = CB', .\ BQ = QB'. And since CA = CB = CB', .-.ZB 'BA is a rt. Z. .-. CQ is _L to BB'. C cannot lie on AB', for if it could, then CC'+C'B would equal CB, which is impossible. Post. 1 Then since AC -i-CB' <AC' -\- C'B', §112 .-. AC-\-CB<AC'-\-C'B'. Ax. 9 .'. AC'-\-C'B<AC'-{-C'B'. Ax. 9 .-. C'B < C'B'. Ax. 6 .-. C ' cannot lie on CQ, for then C 'B would equal C 'B'. § 150 C'cannot lie above CQ (Fig. 1), for C"i>", which < C'P + P^', would be less than C'B, which equals C'P + PB'. .'. C must lie below CQ, as in Eig. 2. .-. A ABC > A ABC, by § 327. Q.e.d. MAXIMA AND MINIMA 267 Proposition V. Theorem 415. Of all polygons with sides all given hut one, the maximum can he inscribed in the semicircle ivhich has the undetermined side for its diameter. c Given ABCDE^ the maximum of polygons with sides AB^ BC^ CDj DEj having the vertices A and E on the line MN. To prove that ABODE can he inscribed in the semicircle having EA for its diameter. Proof. From any vertex, as C, draw CA and CE. The A ACE must be the maximum of all A having the sides CA and CE, and the third side on MN ; otherwise, by increas- ing or diminishing the Z EC A, keeping the lengths of the sides CA and CE unchanged, but sliding the extremities A and E along the line MN, we could increase the A A CE, while the rest of the polygon would remain unchanged; and therefore we could increase the polygon. But this is contrary to the hypothesis that the polygon is the maximum polygon. Hence the A ACE is the maximum of triangles that have the sides CA and CE. Therefore the Z. A CE is a right angle. § 413 Therefore C lies on the semicircle having EA for its diameter. § 215 Hence every vertex lies on this semicircle. That is, the maximum polygon can be inscribed in the semi- circle having the undetermined side for its diameter. q.e.d. 268 APPENDIX TO PLANE GEOMETRY Proposition VI. Theorem 416. Of all polygons ivitJi given sides, one that can he inscribed in a circle is the maximum. Given the polygon ABCDE inscribed in a circle, and the polygon A^B'C'D'E' which has its sides equal respectively to the sides of ABCDE ^ but which cannot be inscribed in a circle. To prove that ABCDE > A'B'C'B'U'. Proof. Draw the diameter AP^ and draw CP and PD. Upon CD' as a base, construct the A C'P'D' congruent to the ACPD, and draw ^'P'. Since, by hypothesis, a O cannot pass through all the vertices of A'B'C'P'D'E', one or both of the parts A'P'D'E', A'B'C'P' cannot be inscribed in a semicircle. Neither A'P'D'E' or A'B'C'P' can be greater than its corre- sponding part. § 415 {Of all polygons with sides all given but one, the maximum can be inscribed in the semicircle which Jias the undetermined side for its diameter.) Therefore one of the parts A'P'D'E', A'B'C'P' must be less than, and the other cannot be greater than, the corresponding part of ABCPDE. .-. ABC PDE> A'B'C'P'D'E'. Take from the two figures the congruent A CPD and C'P'B'. Then ABCDE >A'B'C'D'E', by Ax. 6. Q.e.d. MAXIMA AND MINIMA Proposition VII. Theorem 269 417. Of isoperwietric polygons of a given niiinhe?' of sides, the maximum is equilateral. Given the polygon ABCDEF, the maximum of isoperimetric polygons of n sides. To prove that the polygon ABCDEF is equilateral. Proof. Draw A C. The A ABC must be the maximum of all the A which are formed upon AC with a perimeter equal to that of A ABC. Otherwise a greater A A PC could be substituted for AABC^ without changing the perimeter of the polygon. But this is inconsistent with the fact that the polygon ABCDEF is given as the maximum polygon. .*. the A ABC is isosceles. § 414 .\AB = BC. Similarly BC = CD, CD = DE, and so on. .'. the polygon ABCDEF is equilateral. q.e.d. 418. Corollary. The maximum of isoperimetric polygons of a given number of sides is a regular polygon. For the maximum polygon is equilateral (§ 417), and can be inscribed in a circle (§ 416). Therefore the maximum polygon is regular (§ 365). 270 APPENDIX TO PLANE GEOMETRY Proposition VIII. Theorem 419. Of isoperwietric regular polygons, that ivhich has the greatest nimiber of sides is the maximimi. Y< Given the regular polygon P of three sides, and the isoperimetric regular polygon P' of four sides. To prove that P'>P. Proof. Draw CX from C to any point X in ^jB. Invert the A AXC and place it in the position XCY, letting X fall at C, C at X, and yl at F. The polygon XBCY is an irregular polygon of four sides, which by construction has the same perimeter as P' and the same area as P. Then the regular polygon P' of four sides is greater than the isoperimetric irregular polygon XBCY of four sides. § 418 That is, a regular polygon of four sides is greater than the isoperimetric regular polygon of three sides. In like manner, it may be shown that P ' is less than the iso- perimetric regular polygon of five sides, and so on. q.e.d. Discussion. We may illustrate this by the case of an equilateral tri- angle and a square, eacji with the perimeter.^. In the triangle the base is I p, the altitude I p Vs, and the area J^p2 Vs, or about 0.048 p^. In the square the base and altitude are each i p, and the area is ^^ p^, or 0.0625 jp^. The area of the polygon is therefore increasing as we increase the number of sides. Since the limit approached by the perimeters is a circle, we may infer that of all isoperimetric plane figures the circle has' the greatest area. MAXIMA AND MINIMA 271 Proposition IX. Theorem 420. Of regular polygons having a given area, that which has the greatest nuviber of sides has the minimum 2Jerimeter. Given the regular polygons P and P' having the same area, P' having the greater number of sides. To prove that the perimeter of P > the perimeter of P'. Proof. Construct the regular polygon P" having the same perimeter as P \ and the same number of sides as P. Denote a side of P by s, and a side of P " by s". §419 Given Ax. 9 §374 Then P'>P". But P = P'. .-. P>P". But P:P" = s^:s"\ .\s>s". Ax. 6 .'. the perimeter of P > the perimeter of P". Ax. 6 . But the perimeter of P ' = the perimeter of P ". Const. .'. the perimeter of P > the perimeter of P', by Ax. 9. q.e. d. Discussion. We may illustrate this, as on page 270, by the case of an equilateral triangle and a square, each with area a^. The side of the square is a, and the perimeter 4 a. The area of the equilateral tri- angle is I s2 Vs. Therefore ^ s^ Vs = a^^ or ^ s</d = a. Now </s = VVs ; hence we have V3 = 1.734-,and v V3= Vl. 73 = 1.3 + • Hence is x 1.3 = a, and s = 1.5 a, and the perimeter of the triangle is 4.5 a. Therefore the perimeter of the square is less than that of the triangle. 272 APPENDIX TO PLANE GEOMETRY EXERCISE 73 Maxima and Minima 1. Of all equivalent parallelograms that have equal bases, the rectangle has the minimum perimeter. 2. Of all equivalent rectangles, the square has the minimum perimeter. 3. Of all triangles that have the same base and the same altitude, the isosceles has the minimum perimeter. 4. Of all triangles that can be inscribed in a given circle, the equilateral is the maximum and has the maximum perimeter. 5. To inscribe in a semicircle the maximum rectangle. 6. Find the area of the maximum triangle inscribed in a semicircle whose radius is 3 in. 7. Of all polygons of a given number of sides that can be inscribed in a given circle, that which is regular has the maximum area and the maximum perimeter. 8. Of all polygons of a given number of sides that can be circumscribed about a given circle, that which is regular has the minimum area and the minimum perimeter. 9. In a given line required to find a point such that the sum of its distances from two given points on the same side of the line shall be the minimum. a How does AP -\- PB compare with A'B? and -^ this with A'X-^ XB ? and this with AX+ XB ? J^ This is the problem of a ray of light from -4 to the mirror CD, and reflected to B. 10. To divide a given line into two 1' segments such that the sum of the squares on these segments shall be the minimum. 11. To divide a given line into two segments such that their product shall be the maximum. EECREATIONS 273 421. Recreations of Geometry. The following simple puzzles and recreations of geometry may serve the double purpose of adding interest to the study of the subject and of leading the student to exercise greater care in his demonstrations. They have long been used for this purpose and are among the best known puzzles of geometry. EXERCISE 74 1. To prove that every triangle is isosceles. Let ABC be a A that is not isosceles. Take CP the bisector of ZACB, and ZP the ± bisector of AB. These lines must meet, as at P, for otherwise c they would be II, which would require CP to be _L j/f\ to AB, and this could only happen if A ABC were Y/ / \x isosceles, which is not the case by hypothesis. >: ^'''''\ -/ From P draw PX ± to BC and PY ± to C^, and ^ ^^^:' | "><^ ^ draw PA and PB. '' ^' Then since ZP is the ± bisector of AB, .-. PA = PB. And since CP is the bisector of ZACB, .-. PX = PY. .-. the rt. A PBX and PA Y are congruent, and BX = AY. But the rt. A PXC and PYC are also congruent, and .-. XC = YC. Adding, we have BX + XC = AY -\- YC, or BC = AC. .-. A ABC is isosceles even though constructed as not isosceles. 2. To prove that part of an angle equals the whole angle. Take a square ABCD, and draw MM'P, the ± bisector of CD. Then Jfilf'P is also the ± bisector of ylB. D M c From B draw any line BX equal to AB. Draw BX and bisect it by the _L NP. Since BX intersects CD, Js to these lines can- not be parallel, and must meet as at P. Draw PA, PB, PC, PX, and PB. Since MP is the ± bisector of CD, PD=PC. Similarly PA = PB, and PD = PX. X'fi^'''' .-. PX = PD = PC. K But BX — BC by construction, and PB is common to A PBX and PBC. .: A PBX is congruent to A PBC, and Z XBP = Z CBP. .: the whole Z XBP equals the part, Z CBP, 274 APPENDIX TO PLANE GEOMETRY 3. To prove that part of an angle equals the whole angle. Take a right triangle ABC, and con- struct upon tlie hypotenuse BC an equi- lateral triangle BCD, as shown. On CD lay off CP equal to CA. Through X, the mid-point of AB, draw PX to meet CB produced at Q. Draw QA. Draw the ± bisectors of QA and QP, as YO and ZO. These must meet at some point because they are ± to two intersecting lines. Draw OQ, OA, OP, and OC. Since is on the ± bisector of QA, .-. OQ = OA. Similarly OQ = OP, and .-. OA = OP. But CA = CP, by construction, and CO = CO. .: AA OC is congruent to A POC, and ZACO = Z PCO. 4. To prove that part of a line equals the whole line. Take a triangle ABC, and draw CP ± to ^-B. From C draw CX, making Z^CX = ZB. Then A ABC and ACX are similar. .-. AABC:AACX=BC^:CX^. Furthermore A ABC: A ACX = AB: AX. .'. BC" " or BC^:AB=CX But A' P CX^ = AB:AX, ^:AX. BC^ = AC'^+ AB^ and CX' AC^ + AX^ f;2 AC'-{-AB"-2AB-AP AC^+AX^-2AX 2AB.AP, 2AX-AP. AP AB AX ^2 AC AB AC ^ AB-2AP = —— + AX-2AP. AX AC' AB AX ACT AX AB, AC'-AB-AX AC'- AB-AX AB AX AB= AX. RECREATIONS 275 5. To show geometrically that 1=0. Take a square that is 8 units on a side, and cut it into three parts, -4, 7?, C, as shown in the right-hand figure. Fit these parts together as in the left-hand figure. Now tlie square is 8 units on a side, and therefore contains 8 x 8, or 64, small squares, while the rec- tangle is 13 units long and 5 units high, and there- fore contains 5 X 13, or 65, small squares. But the two figures are each made up of J. + J5+ C (Ax.ll), and therefore are equal (Ax.8). 64 we have 1 z= (Ax. 2). + _4__,__(-_. 65 = 64, and by subtracting 6. To prove that any point on a line bisects it. Take any point P on AB. q On AB construct an isosceles A ABC, having /*\ AC=BC; and draw PC. / \\ Then in A APC and PBC, we have / \ \ ZA=ZB, § 74 / \ \ AC=BC, Const. / \ \ and PC = PC. Iden. "^ J' ^ Three independent parts (that is, not merely the three angles) of one triangle are respectively equal to three parts of the other, and the tri- angles are congruent ; therefore AP = BP (§ 67). 7. To prove that it is possible to let fall two perpendiculars to a line from an external point. Take two intersecting © with centers and (7. Let one point of intersection be P, and draw the diameters PA and PD. Draw AD cutting the circumferences at B and C. Then draw PB and PC. Since Z PC A is inscribed in a semicircle, it is a right angle. In the same way, since ZDBP is inscribed in a semicircle, it also is a right angle. .-. PB and PC are both X to AD. 276 APPENDIX TO PLANE GEOMETRY 8. To prove that if two opposite sides of a quadrilateral are equal the figure is an isosceles tra]oezoid. Given the quadrilateral ABCD, with BC = DA. To prove that ^J5 is |I to DC. Draw MO and NO, the ± bisectors of AB and CD, to meet at 0. li AB and DC are parallel, the proposition is already proved. U AB and DC are not parallel, then MO and NO will meet at 0, either inside or outside the figure. Let be supposed to be inside the figure. Draw OA, OB, OC, OD. Then since OM is the ± bisector of AB, .-. OA = OB. Similarly and Also, and Similarly and OD = OC. But DA is given equal to BC. .'. AAOD is congruent to A BOC, ZDOA = ZBOC. rt. ii OCN and ODN are congruent, ZNOD = ZCON. rt. A AMO and BMO are congruent, ZAOM = ZMOB. .: Z NOD + Z DO A -\-ZAOM=Z CON + Z BOC + Z MOB, or Z NOM = Z MON = a st. Z. Therefore the line MON is a straight line, and hence AB is II to DC. D ^^^, If the point is outside the quadrilateral, as In the second figure, the proof is substantially the same. For it can be easily shown that A/ ZDON-ZDOA-ZAOM 1/ = Z NOC - Z BOC - Z MOB, ^ which is possible only if ZDON=ZDOM, or if ON lies along OM. But that the proposition is not true is evident from the third figure, in which BC = DA, but ^Z^ is not II to DC. O 'N \ M HISTORY OF GEOMETRY 277 422. History of Geometry. The geometry of very ancient peoples was largely the mensuration of simple areas and volumes such as is taught to children in elementary arithmetic to-day. They learned how to find the area of a rectangle, and in the oldest mathematical records that we have there is some discussion of triangles and of the volumes of solids. The earliest documents that we have, relating to geometry, come to us from Babylon and Egypt. Those from Babylon were written about 2000 b.c. on small clay tablets, some of them about the size of the hand, these tablets afterwards having been baked in the sun. They show that the Baby- lonians of that period knew something of land measures, and perhaps had advanced far enough to compute the area of a trapezoid. For the mensuration of the circle they later used, as did the early Hebrews, the value tt = 3. The first definite knowledge that we have of Egyptian math- ematics comes to us from a manuscript copied on papyrus, a kind of paper used about the Mediterranean in early times. This copy was made by one Aah-mesu (The Moon-born), com- monly called Ahmes, who probably flourished about 1700 b.c. The original from which he copied, written about 2300 b.c, has been lost, but the papyrus of Ahmes, written nearly four thousand years ago, is still preserved and is now in the British Museum. In this manuscript, which is devoted chiefly to frac- tions and to a crude algebra, is found some work on mensu- ration. Among the curious rules are the incorrect ones that the area of an isosceles triangle equals half the product of the base and one of the equal sides; and that the area of a trapezoid having bases &, h\ and nonparallel sides each equal to (X, is ^a(b-\-b'). One noteworthy advance appears however. Ahmes gives a rule for finding the area of a circle, substan- tially as follows : Multiply the square on the radius by (-V-)^ which is equivalent to taking for ir the value 3.1605. Long before the time of Ahmes, however, Egypt had a good working 278 APPENDIX TO PLANE GEOMETRY knowledge of practical geometry, as witness the building of the pyramids, the laying out of temples, and the digging of irrigation canals. From Egypt and possibly from Babylon geometry passed to the shores of Asia Minor and Greece. The scientific study of the subject begins with Thales, one of the Seven Wise Men of the Grecian civilization. Born at Miletus about 640 b.c, he died at Athens in 548 b.c. . He spent his early manhood as a merchant, accumulating the wealth that enabled him to spend his later years in study. He visited Egypt and is said to have learned such elements of geometry as were known there. He founded a school of mathematics and philosophy at Miletus, known as the Ionic School. How elementary the knowledge of geometry then was, may be understood from the fact that tradition attributes only about four propositions to Thales, substantially those given in §§ 60, 72, 74, and 215 of this book. The greatest pupil of Thales, and one of the most remark- able men of antiquity, was Pythagoras. Born probably on the island of Samos, just off the coast of Asia Minor, about the year 580 b.c, Pythagoras set forth as a young man to travel. He went to Miletus and studied under Thales, probably spent several years in study in Egypt, very likely went to Babylon, and possibly went even to India, since tradition asserts this and the nature of his work in mathematics confirms it. In later life he went to southern Italy, and there, at Crotona, in the southeastern part of the peninsula, he founded a school and established a secret society to propagate his doctrines. In geometry he is said to have been the first to demonstrate the proposition that the square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides (§ 337). The proposition was known before his time, at any rate for special cases, but he seems to have been the first to prove it. To him or to his school seems also to have been due the construction of the regular pentagon (§§ 397, 398) HISTORY OF GEOMETRY 279 and of the five regular polyhedrons. The construction of the regular pentagon requires the dividing of a line in extreme and mean ratio (§ 311), and this problem is commonly assigned to the Pythagoreans, although it played an important part in Plato's school. Pythagoras is also said to have known that six equilateral triangles, three regular hexagons, or four squares, can be placed about a point so as just to fill the 360°, but that no other regular polygons can be so placed. To his school is also due the proof that the sum of the angles of a triangle equals two right angles (§ 107), and the construction of at least one star-polygon, the star-pentagon, which became the badge of his fraternity. For two centuries after Pythagoras geometry passed through a period of discovery of propositions. The state of the science may be seen from the fact that (Enopides of Chios, who flourished about 465 b.c, showed how to let fall a perpendicu- lar to a line (§ 227), and how to construct an angle equal to a given angle (§ 232). A few years later, about 440 b.c, Hippoc- rates of Chios wrote the first Greek textbook on mathematics. He knew that the areas of circles are proportional to the squares on their radii, but was ignorant of the fact that equal central angles or equal inscribed angles intercept equal arcs. About 430 B.C. Antiphon and Bryson, two Greek teachers, worked on the mensuration of the circle. The former attempted to find the area by doubling the number of sides of a regular inscribed polygon, and the latter by doing the same for both in- scribed and circumscribed polygons. They thus substantially exhausted the area between the circle and the polygon, and hence this method was known as the Method of Exhaustions. During this period the great philosophic school of Plato (429-348 B.C.) flourished at Athens, and to this school is due the first systematic attempt to create exact definitions, axioms, and postulates, and to distinguish between elementary and higher geometry. At this time elementary geometry became 280 APPENDIX TO PLANE GEOMETRY limited to the use of the compasses and the unmarked straight edge, which took from this domain the possibility of con- structing a square equivalent to a given circle (" squaring the circle"), of trisecting any given angle, and of constructing a cube with twice the volume of a given cube (" duplicating the cube"), these being the three most famous problems of antiquity. Plato and his school were interested in the so-called Pythagorean numbers, numbers that represent the three sides of a right triangle. Pythagoras had already given a rule to the effect that i(m''-{-lf = 7n^ + i(m^-iy. The school of Plato found that [(^my-\-iy = m^-i-[(^my-lf. By giving various values to vi, different numbers will be found such that the sum of the squares of two of them is equal to the square of the third. The first great textbook on geometry, and the most famous one that has ever appeared, was written by Euclid, who taught mathematics in the great university at Alexandria, Egypt, about 300 B.C. Alexandria was then practically a Greek city, having been named in honor of Alexander the Great, and being ruled by the Greeks. Euclid's work is known as the "Elements," and, as was the case with all ancient works, the leading divisions were called books, as is seen in the Bible and in such Latin writers as Caesar and Vergil. This is why we speak of the various books of geometry to-day. In this work Euclid placed all the leading propositions of plane geometry as then known, and arranged them in a logical order. Most subsequent geometries of any im- portance since his time have been based upon Euclid, improving the sequence, symbols, and wording as occasion demanded. Euclid did not give much solid geometry because not much was known then. It was to Archimedes (287-212 b.c), a famous mathematician of Syracuse, on the island of Sicily, that some of the most important propositions of solid geometry are due, particularly those relating to the sphere and cylinder. HISTORY OF GEOMETRY 281 He also showed how to find the approximate value of tt by a method similar to the one we teach to-day (§ 404), proving that the real value lies between 3} and 3}^. Tradition says that the sphere and cylinder were engraved upon his tomb. The Greeks contributed little more to elementary geometry, although Apollonius of Perga, who taught at Alexandria be- tween 250 and 200 b.c, wrote extensively on conic sections; and Heron of Alexandria, about the beginning of the Christian era, showed that the area of a triangle whose sides are a, b, c, equals V6'(.s- — a) (s — b) (s — c), where s = ^(a-\-b^-c) (see p. 211). The East did little for geometry, although contributing considerably to algebra. The first great Hindu writer was Aryabhatta, who was born in 476 a.d. He gave the very close approximation for tt, expressed in modern notation as 3.1416. The Arabs, about the time of the Arabian Nights Tales (800 A.D.), did much for mathematics, translating the Greek authors into their own language and also bringing learning from India. Indeed, it is to them that modern Europe owes its first knowledge of Euclid. They contributed nothing of importance to geometry, however. Euclid was translated from the Arabic into Latin in the twelfth century, Greek manuscripts not being then at hand, or being neglected because of ignorance of the language. The leading translators were Athelhard of Bath (1120), an English monk who had learned Arabic in Spain or in Egypt ; Gerhard of Cremona, an Italian monk ; and Johannes Campanus, chap- lain to Pope Urban IV. In the Middle Ages in Europe nothing worthy of note was added to the geometry of the Greeks. The first edition of Euclid was printed in Latin in 1482, the first one in English appearing in 1570. Our symbols are modern, -|- and — first appearing in a German work in 1489; = in Recorde's "Whet- stone of Witte" in 1557; > and < in the works of Harriot (1560-1621); and x in a publication by Oughtred (1574-1660). 282 APPENDIX TO PLANE GEOMETRY 423. Notation used in Formulas. Following the general cus- tom, small letters represent numerical values, large letters rep- resent points. The following abbreviations have been used in this book as consistently as the circumstances would allow, the context telling which abbreviation is intended : a = area, apothem I = length. a, b, c = sides oiAABC. m = median, a' = projection of a. 2^ — perimeter. b, h' = bases. ^ r = radius. c = circumference. s = semiperimeter of A, d = diameter, diagonal. 1 (a + Z» -f c). h = height, altitude. ir = 3.1416, or about 3}. 424. Formulas for Line Values. The following are the most important formulas in line values : Right triangle, a' -i-b'' = c" (§ 337). Any triangle, a^ -^ b'' ± 2 aV = c" {^% 341, 342). Circle, c = 2 irr = ird (§ 385). Radius of circle, v — c-^2'Tr. Equilateral triangle, ^^ = \b y 3. Diagonal of square, d = b V2 (§ 339). Side of square, b = Va. 425. Areas of Plane Figures. The following are the formulas for the areas of the most important j^lane figures : Rectangle, bh (§ 320). Square, b'' (§ 320). Parallelogram, bh (§ 322). Triangle, ^^bh(^S25),^s(s-a)(s-b){s-c). Equilateral triangle, ^ b'^ v 3. Trapezoid, V^(^ + ^') (§^29). Regular polygon, ^ ap (§ 386). Circle, ^-rc = 7rr2(§§388, 389). INDEX Page Acute angle 16 Acute triangfe 26 Adjacent angles 7 Alternation, proportion by .152 Altitude 59 Analytic proof . . 80, 140, 141 Angle 6 acute 16 at center of regular polygon 227 central 93 complement of .... 18 conjugate of 18 exterior 51 inscribed 115 measure of 18 oblique 16 obtuse 16 plane 6 reentrant . . . ^ . . 08 reflex 16 right 7, 16 sides of 6 size of 6, 17 straight 16 supplement of 18 vertex of 6 Angles, adjacent 7 alternate-interior . . . 47 complementary .... 18 conjugate 18 corresponding 26 equal 6 Page Angles, exterior . . , . 47, 51 exterior-interior .... 47 generation of 17 interior 47, 51 made by a transversal . . 47 of a polygon 68 of a triangle 7 supplementary . . . . 18 vertical 18 Antecedents 151 Apothem 227 Arc 7,93 major 93 minor 93 Area of circle 115 of irregular polygon . . 199 of surface 191 Attack, methods of . . 140, 145 Axiom 21 Axioms, list of 22 Axis of symmetry .... 261 Base . 7, 32, 59 Bisector 6, 74 Broken line 5 Center of circle 7 of regular polygon . . . 227 of symmetry 261 Central angle 93 Chord 95 Circle 7,93 283 284 INDEX Page Circle, arc of . . . . . 7, 93 area of ... . . . .115 as a limit . . . . 114, 237 as a locus . . . ... 93 center of . . . ... 7 central angle of . ... 93 chord of . . . ... 95 circumference of . ... 7 circumscribed . . . . .114 diameter of . . . . 7, 93 inscribed . . . ... 114 radius of . . . . . 7, 93 secant to . . , . 102, 177 sector of . . . . . . 115 segment of . . . . . .115 tangent to . . . . . .102 Circles, concentric . . . .104 escribed .... ... 137 tangent .... . . .107 Circumcenter . . . . . 78, 136 Circumference . . ... 7 Circumscribed circle . . .114 Circumscribed polygor 1 . . .114 Commensurable magn tudes . 112 Common measure . . .112 Common tangents . . .109 Complement . . . ... 18 Composition, proportic m by .153 Concave polygon . . ... 68 Concentric circles . ... 104 Concurrent lines . . .... 77 Congruent .... . . 26,68 Conjugate .... ... 18 Consequents . . . . . .151 Constant .... . . .114 Continued proportion . . .151 Continuity, principle c )f . . 125 Converse propositions . . 35, 95 Converse theorems, la w of .95 Convex polygon . . . , . 68 Page Corollary 21 Corresponding angles ... 26 Corresponding lines .... 165 Corresponding sides . . 26, 165 Curve 5 Curvilinear figure .... 6 Decagon 68 Degree .... .^ ... 18 Determinate cases .... 140 Diagonal 59, 68 Diameter 7, 93 Difference of magnitudes . . 17 Dimensions 2 Discussion of a problem 126, 140 Distance 42 Division, harmonic .... 161 proportion by 154 Dodecagon ....... 68 Drawing figures . . . 8, 29, 84 Equal angles 6 Equal lines 5 Equiangular polygon ... 68 Equiangular triangle ... 26 Equilateral polygon .... 68 Equilateral triangle .... 26 Equivalent figures . . . . 191 Escribed circles 137 Excenter 137 Exterior angles . . . . 47, 51 Extreme and mean ratio . . 184 Extremes ....... 151 Figure 4 curvilinear 5 geometric ...... 4 plane 4 rectilinear 6 symmetric 261 INDEX 285 Figures, equivalent . isoperimetric . . symmetric . . . Foot of perpendicular Formulas .... Fourth proportional . Generation of angles of magnitudes . . Geometric figure . . Geometry . . . .' History of . . . Harmonic division . Heptagon . . . . Hexagon . . . . History of Geometry Homologous angles . Homologous lines Homologous sides Hypotenuse . . . Hypothesis . . . . Impossible cases Incenter . . . Incommensurable magnitudes Incommensurable ratio Indeterminate cases . Indirect proof . Inscribed angle . . Inscribed circle . Inscribed polygon Instruments Interior angles . . Inversion, proportion Isoperimetric polygons Isosceles trapezoid . Isosceles triangle . . by Limit Page page . 191 Limits, principle of ... . 115 . 265 Line 3,5 . 261 broken 5 7 curve 5 . 282 of centers 107 . 151 segments of .... 5, 161 straight 5 . 17 Lines, concurrent . . . .77 4,17 corresponding. , .• . , 165 4 equal 5 4 oblique 16 . 277 parallel 46 perpendicular 7 product of 194 transversal of 47 Loci, solutions by .... 143 Locus 73 proof of 74 Magnitudes 3 bisectors of 6 commensurable .... 112 constant 114 differences of 17 generation of . . . . 4, 17 incommensurable . . .112 sums of 17 variable 114 Maximum 265 Mean proportional .... 151 Means 151 Measure 112 . 8 angle 18 47, 51 common 112 . 153 numerical .... 112, 117 . 265 Median 77 . 59 Methods of attack . . 140, 145 . 26 of proof. . 35,77,80,8.3,84 Minimum 265 114, 237 Multiple 112 161 277 26 165 26 42 30 140 137 112 113 140 83 115 114 114 286 INDEX Page Nature of proof 25 of solution 126 Negative quantities .... 125 Nonagon 68 Numerical measure . . 112, 117 Oblique angle 16 Oblique lines 16 Obtuse angle 16 Obtuse triangle 26 Octagon . . 68 Optical illusions 15 Parallel lines 46 Parallelogram 59 Pentagon 68 Pentadecagon 246 Perigon 18 Perimeter 7, 68 Perpendicular 7 Perpendicular bisector ... 74 Pi(7r) 238 value of 249 Plane 3 angle 6 geometry 4 Point 3 of contact .... 102, 107 Polygon 68 angles of 68 apothem of regular . . .227 area of 191, 199 center of regular .... 227 circumcenter of ... . 136 circumscribed 114 concave 68 convex 68 diagonal of .... 59, 68 equiangular 68 equilateral 68 Page Polygon, incenter of . . . 137 inscribed 114 perimeter of 68 radius of regular .... 227 regular . 68, 227 sides of 68 vertices of 68 Polygons, classified . . 68, 114 congruent 68 isoperimetric 265 mutually equiangular . . 68 mutually equilateral . . 68 similar 165 Positive quantities .... 125 Postulate . 21 of parallels 46 Postulates, list of .... 23 Principle of continuity . . . 125 of limits 115 Problem 21, 126 how to attack . . . 140, 145 Product of lines 194 Projection 205 Proof, methods of 35, 77, 80, 83, 84 nature of 25 necessity for 15 Proportion 151 continued 151 nature of quantities in a .155 Proportional, fourth . . . .151 mean 151 reciprocally 177 third 151 Proposition 21 Pythagorean theorem . . . 204 Quadrilateral 59, 68 Quadrilaterals classified . . 59 Radius 7, 93 of regular polygon . . . 227 INDEX 287 Page Ratio 112 extreme and mean . . .184 incommensurable . . .113 of similitude 165 Recreations of Geometry . . 273 Rectangle 59 Rectilinear figure 5 Reductio ad absurdum ... 83 Reentrant angle 68 Reflex angle 16 Regular polygon . . . .68, 227 Rhomboid 59 Rhombus 59 Right angle 7, 16 Right triangle 26 Scalene triangle 26 Secant 102, 177 Sector 115 Segment of a circle . . . .115 of a line 5, 161 Semicircle 93, 115 Sides, corresponding ... 26 of angle 6 of triangle 7 of polygon . . ... . 68 Similar parts of circles . . . 239 Similar polygons 165 Similitude, ratio of ... . 165 Size of angle 6, 17 Solid 2 Solid geometry 4 Solution, nature of ... . 126 Square 26 Straight angle 16 Straight line 5 Subtend 93, 95 Sum of magnitudes . . . . 17 Superposition 35 Supplement 18 Page Surface 3, 191 Symmetric figures .... 261 Symmetry 261 Synthetic proof . . 35, 77, 140 Tangent .... 102, 107, 109 Tangent circles 107 Terms of a proportion . . .151 Theorem . 21 Third proportional . . . .151 Transversal 47 Trapezium 59 Trapezoid 59 Triangle 7, 68 acute 26 altitude of 59 angles of 7 base of 7, 59 circumcenter of . . . 78, 136 equiangular 26 equilateral 26 excenter of 137 incenter of 78, 137 isosceles 26 obtuse 26 right 26 scalene 26 sides of 7 vertices of 7 Triangles classified .... 26 Unit of measure 112 of surface 191 Variable 114 Vertex of angle 6 of isosceles triangle . . 59 Vertical angles 18 Vertices of a polygon ... 68 of a triangle 7 RETURN CIRCULATION DEPARTMENT TOaii^ 202 Main Library 642-3403 LOAN PERIOD 1 HOME USE 2 3 4 5 6 . 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